INTRO CALCULUS
(DIFFERENTIATION and APPLICATIONS)
THE ENTIRE COURSE IN ONE BOOK
September 2014 edition
This book covers the topics on a
typical midterm and final exam.
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TABLE OF CONTENTS
Formulas to Memorize .......................................................................................................1
Required Theorem Proofs to Memorize .............................................................................3
Lesson 1: Skills Review ....................................................................................................5
Lesson 2: Limits .............................................................................................................31
Homework and Limits Practise Problems ..............................................................68
Solutions to Limits Practise Problems ...................................................................70
Lesson 3: Continuity .......................................................................................................79
Homework and Continuity Practise Problems ........................................................90
Solutions to Continuity Practise Problems .............................................................93
Lesson 4: The Definition of Derivative ............................................................................96
Homework and Definition of Derivative Practise Problems .................................106
Solutions to Definition of Derivative Practise Problems ...................................... 107
Lesson 5: The Differentiation Rules .............................................................................. 110
Hints on the Required Theorem Proofs ................................................................ 152
Homework and Differentiation Rules Practise Problems ..................................... 160
Solutions to Differentiation Rules Practise Problems .......................................... 164
Lesson 6: Implicit Differentiation ................................................................................. 173
Homework and Implicit Differentiation Practise Problems ..................................183
Solutions to Implicit Differentiation Practise Problems ....................................... 185
Lesson 7: Related Rates ................................................................................................ 190
Steps to Solving Related Rates Problems ............................................................ 190
Homework and Related Rates Practise Problems ................................................ 201
Solutions to Related Rates Practise Problems ..................................................... 203
The Midterm Exam usually includes Lessons 1 to 7.
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TABLE OF CONTENTS (continued)
Lesson 8: Log and Exponential Derivatives ..................................................................209
Homework and Log and Exponential Derivatives Practise Problems ...................234
Solutions to Log and Exponential Derivatives Practise Problems ........................238
Lesson 9: Curve-Sketching ...........................................................................................248
Steps to Curve-Sketching ....................................................................................248
Hints on the Required Theorem Proofs ................................................................283
Homework and Curve-Sketching Practise Problems ............................................287
Solutions to Curve-Sketching Practise Problems .................................................293
Lesson 10: Max/Min Word Problems ............................................................................308
Steps to Solving Max/Min Word Problems...........................................................308
Homework and Max/Min Word Problems Practise Problems ...............................329
Solutions to Max/Min Word Problems Practise Problems ....................................332
Lesson 11: Antiderivatives (Integrals) .........................................................................340
Homework and Antiderivatives (Integrals) Practise Problems ............................374
Solutions to Antiderivatives (Integrals) Practise Problems .................................378
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Intro Calculus
1
Formulas and Definitions to Memorize
The Definition of Continuity: f  x  is continuous at x=a if and only if lim f  x   f  a  .
xa
The Definition of Derivative:
 x   nx
n
The Power Rule:
The Quotient Rule:
The Chain Rule:
h0
n 1
 f  g  
The Product Rule:
f  x  h  f  x 
h
f   x   lim
f  g  f g
 T  T  B  T B
  
B2
 B
 f  u    f   u  u
 u   nu
n
The Chain Rule Version of Power Rule:
n 1
 u
Derivatives of Trigonometric Functions:
 sin u   cos u  u
 tan u   sec2 u  u
 sec u   sec u tan u  u
 cos u    sin u  u
 cot u    csc2 u  u
 csc u    csc u cot u  u
Derivatives of Exponential and Logarithmic Functions:
 e   e
u
u
 a   a
u
u
u
u
 u
 ln u  
 u  ln a
 log a u  
 f   a  
1
Derivative of an Inverse Function:
Fundamental Theorem of Calculus:
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u
 u  ln a
1
f   f 1  a  

 u f t dt   f u  u
 
   
a

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2
Intro Calculus
Antiderivative Formulas:
 K dx  Kx  C
x




n
dx 
1
n1
x n1  C
1
dx  ln x  C
x
e
x
dx  e x  C ,
x
 a dx 
so
e
ax
dx 
1 ax
e C
a
e.g.
e
5x
dx 
1 5x
e C
5
so
 sin ax dx   1a cos ax  C
e.g.
 sin  3x  dx   13 cos  3x   C
so
 cos ax dx  1a sin ax  C
e.g.
 cos 7 x  dx  71 sin 7 x   C
ax
C
ln a
 sin x dx   cos x  C ,
 cos x dx  sin x  C ,
2
x dx  tan x  C ,
since
 tan x   sec2 x
2
x dx   cot x  C ,
since
 cot x    csc2 x
 sec x tan x dx  sec x  C ,
since
 sec x   sec x tan x
 csc x cot x dx   csc x  C ,
since
 csc x    csc x cot x
 sec
 csc
Trigonometric Values to Memorize
0
© Grant Skene for
sin
0
cos
1
tan
0
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4
3
1
2
2
2
3
2
1
3
2
2
3
2
1
2
1
3
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2
1
0
undefined
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Intro Calculus
3
Required Theorem Proofs to Memorize
(See pages 152 to 155 and pages 283 to 286
for hints to help understand these proofs.)
Theorem 1: (Differentiable Functions are Continuous)
(a) Prove: If a function f is differentiable at point a, then it is
continuous at a.
(b) Is it true that if f is continuous at a it is also differentiable at
a? Justify your answer.
Theorem 4: (The Product Rule)
Given that f and g are differentiable functions, prove:
 f  x  g  x   f   x  g  x   f  x  g  x  .
Proof:
f  x  h g  x  h  f  x  g  x 
 f  x  g  x   hlim
0
h
Proof:
(a) If f is differentiable at a, then f   a  exists where
f   a   lim
h0
f  a  h  f  a 
.
h
 lim
lim f  a  h   lim f  a  h 
h0
h0
 lim  f  a  h   f  a   f  a  
h0
 f  a  h  f  a 

 lim 
 h  f  a 
h0 
h

f a  h  f a 
 lim h  lim f  a 
h0
h0
h0
h
 f a  0  f a
 lim
h
 f  x  h g  x  h  g  x  h f  x  g  x  h f  x   f  x  g  x  
 lim 


h0 
h
h
 f  x  h  f  x 
 lim 
h0 
 lim
h
f  x  h  f  x 
h
h0
 g  x  h  f  x  
g  x  h  g  x  
 lim g  x  h   lim f  x   lim
h0
 f   x  g  x   f  x  g   x  proven
h0

h
g  x  h  g  x 
h0
h
(Note that lim g  x  h  g  x  since g is continuous;
h0
differentiable functions are continuous.)
 f a
We have proven lim f  a  h  f  a  meaning f is
h0
continuous at a.
(b) FALSE. A function can be continuous at a but not be
differentiable at a. For example, f  x   x is continuous at 0
but it is not differentiable at 0.
Theorem 5: (The Derivative of sin x.)
d
Prove:  sin x   cos x (i.e.
sin x  cos x ).
dx
Proof:
Theorem 2: (The Constant Multiple Rule)
Given that c is a constant and f is a differentiable function, prove:
 c  f  x   c  f   x  .
Proof:
f  x  h g  x  h  g  x  h f  x   g  x  h f  x   f  x  g  x 
h0
c  f  x  h  c  f  x 
h
f  x  h  f  x 
 c  lim
h0
h
 c  f   x  proven
 c  f  x    hlim
0
Theorem 3: (The Sum Rule)
Given that f and g are differentiable functions, prove:
sin  x  h   sin x
h
sin x cos h  cos x sin h  sin x
 lim
h0
h
sin x cos h  sin x  cos x sin h
 lim
h0
h
 sin x cos h  sin x cos x sin h 
 lim 


h0 
h
h
 sin x  
lim
h0
cos h  1
sin h 

 lim sin x 
 cos x 
h0 
h
h 
cos h  1
sin h
 lim sin x  lim
 lim cos x  lim
h0
h0
h0
h0 h
h
 sin x  0  cos x  1
 cos x proven
 f  x   g  x   f   x   g  x  .
Proof:
 f  x   g  x    hlim
0
f  x  h  g  x  h   f  x   g  x 
h
f  x  h  f  x   g  x  h  g  x 
 lim
h0
h
 f  x  h  f  x  g  x  h  g  x  
 lim 


h0 
h
h

f  x  h  f  x 
g  x  h  g  x 
 lim
 lim
h0
h0
h
h
 f   x   g   x  proven
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4
Intro Calculus
The Mean-Value Theorem
Given the function f which is continuous on the closed
interval [a, b] and differentiable on the open interval (a, b),
then there exists at least one number c in (a, b) such that:
f c  
f  b  f  a 
ba
Theorem 6:
If f   x   0 for all x in an interval (a, b), then f is constant on
(a, b).
Proof:
Let x1 and x 2 be any two numbers in (a, b) such that x1  x2
.
Clearly f  x  is differentiable and continuous on  x1 , x2  .
Therefore, the Mean-Value Theorem applies. There exists at
least one number c in  x1 , x2  such that:
f c  
f  x2   f  x1 
x2  x1
But, f   c   0 since f   x   0 on  x1 , x2  . Therefore:
0
f  x2   f  x1 
 f  x2   f  x1   0  f  x2   f  x1 
x2  x1
The only way that f  x2   f  x1  for any two numbers x1 and
x 2 in (a, b) is if f is constant on (a, b). Proven
Theorem 7: (Increasing Functions)
(a) Suppose f is continuous on [a, b] and differentiable on (a, b).
Prove:
If f   x   0 for all x in (a, b), then f is increasing on [a, b].
(b) Is it true that if f is increasing on [a, b], then f   x   0 for
all x in (a, b)? Justify your answer.
Proof:
(a) Let x1 and x 2 be any two numbers in [a, b] such that
x1  x2 .
Clearly f  x  is differentiable and continuous on  x1 , x2  .
Therefore, the Mean-Value Theorem applies. There exists at
least one number c in  x1 , x2  such that:
f c  
f  x2   f  x1 
x2  x1
But, f   c   0 since f   x   0 on  x1 , x2  . Therefore:
f  x2   f  x1 
 0  f  x2   f  x1   0  f  x2   f  x1 
x2  x1
Since f  x2   f  x1  , for any x1  x2 in [a, b], then f is
increasing on [a, b]. Proven
(b) No, it is not true that if f is increasing on [a, b], then
3
f   x   0 for all x in (a, b). For example, f  x   x is
increasing on  ,   , but f   x   0 on  ,   ,
f   x   3x 2  0 when x=0 ( f   x   0 for all other values of x
though). i.e., Even though f  x   x 3 is increasing, f   x   0
Definitions of Increasing and Decreasing Functions:
Let x1 and x 2 be any two numbers in an interval I such that
x1  x2 .
If f  x2   f  x1  on I, then f is an increasing function.
If f  x2   f  x1  on I, then f is a decreasing function.
at x=0.
Theorem 8: (Decreasing Functions)
(a) Suppose f is continuous on [a, b] and differentiable on (a, b).
Prove:
If f   x   0 for all x in (a, b), then f is decreasing on [a, b].
(b) Is it true that if f is decreasing on [a, b], then f   x   0 for
all x in (a, b)? Justify your answer.
Proof:
(a) Let x1 and x 2 be any two numbers in [a, b] such that
x1  x2 .
Clearly f  x  is differentiable and continuous on  x1 , x2  .
Therefore, the Mean-Value Theorem applies. There exists at
least one number c in  x1 , x2  such that:
f c  
f  x2   f  x1 
x2  x1
But, f   c   0 since f   x   0 on  x1 , x2  . Therefore:
f  x2   f  x1 
 0  f  x2   f  x1   0  f  x2   f  x1 
x2  x1
Since f  x2   f  x1  , for any x1  x2 in [a, b], then f is
decreasing on [a, b]. Proven
(b) No, it is not true that if f is decreasing on [a, b], then
3
f   x   0 for all x in (a, b). For example, f  x    x is
decreasing on  ,   , but f   x   0 on  ,   ,
f   x   3x 2  0 when x=0 ( f   x   0 for all other values of x
3
though). i.e., Even though f  x    x is decreasing,
f   x   0 at x=0.
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(Intro Calculus) LESSON 1: SKILLS REVIEW
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Lesson 1: Skills Review
Lecture Problems:
(Each of the questions below will be discussed and solved in the lecture that follows. Your
homework for this lesson is to be able to answer all of the questions below without any help.)
Make a sign diagram for the following functions.
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1.
(a)
x2  4 x  4
f x 
x2  9
(b)
 x  3 x  1
f x 
2
 x  2  x  5
(c)
18  2x  x  1
f x 
 x  3 x  2
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3
2
2.
2
2
4
Find the domain of the following functions.
f x 
x 1
x 2
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(a)
(c)
f x 
x2  4
x 2  2x  3
(b)
f x 
x 3
(d)
f x 
x 2  5x  6
x2  4 x  4
3.
Find the domain and range of f  x   9  x 2 .
4.
Solve for x in the equations below:
(a)
 3.4 2 x
7
(d) x 3 5  8
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(Intro Calculus) LESSON 1: SKILLS REVIEW
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(Intro Calculus) LESSON 1: SKILLS REVIEW
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(Intro Calculus) LESSON 2: LIMITS
31
Lesson 2: Limits
Memorize these two trig limits: lim
h0
sin h
1
h
and
lim
h0
cos h  1
0
h
Lecture Problems:
(Each of the questions below will be discussed and solved in the lecture that follows.)
3.
5.
7.
x  1
lim
x 3
2.
4  x2  7
4 x 2  5 x  21
lim
x  1
4.
x 2  3x  4
5x 2  9 x  4
6.
x2  9
x2  x  2
lim
x  2
8.
x  7 x  18
2
lim
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9.
x2  x  2
x 1
lim
11.
x   2
4 x 3  3x  2
2
x   3 x  5 x  1
lim
 2 x  3  x 2  6 x  5
10.
12.
2
13.
lim
x4  x3  6
x 
x2
lim
x4  2
x  2
x 2
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1.
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For each of questions 1 to 14, find the value of the limit, if it exists. If it does not exist, is
it infinity, negative infinity, or neither? Justify your answers.
14.
lim
x2
lim
x 5
x 2
3 x
x  10 x  25
lim
x 6
2
x 6
2x 3  x  2
lim
x
1  3x 3
lim
x  
5 x  21
4 x 2  3x  7
1  3 x  4 x 2  5
2
x  
5x  4 
lim
15.
Which limits in questions 1 to 14 above indicate the existence of a Vertical or
Horizontal Asymptote?
16.
Find the following limits.
(a)
17.
lim
sin  2 x   7 x
3x
x 0
(b)
lim
x 0
sin  2 x 
sin  5 x 
(c)
lim
x 0
tan  4 x 
2x
Use the Squeeze Theorem to solve the following limits.
(a)
lim
x 0
1
x 4 sin2  
x
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 3 
lim x 5 cos  3 
x 0
x 
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LESSON 2: LIMITS (Intro Calculus)
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© 1997-2011 Grant Skene for
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(Intro Calculus) LESSON 2: LIMITS
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LESSON 2: LIMITS (Intro Calculus)
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(Intro Calculus) LESSON 2: LIMITS
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LESSON 2: LIMITS (Intro Calculus)
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© 1997-2011 Grant Skene for
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(Intro Calculus) LESSON 2: LIMITS
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LESSON 2: LIMITS (Intro Calculus)
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(Intro Calculus) LESSON 2: LIMITS
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LESSON 2: LIMITS (Intro Calculus)
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(Intro Calculus) LESSON 2: LIMITS
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LESSON 2: LIMITS (Intro Calculus)
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(Intro Calculus) LESSON 2: LIMITS
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LESSON 2: LIMITS (Intro Calculus)
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(Intro Calculus) LESSON 2: LIMITS
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LESSON 2: LIMITS (Intro Calculus)
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(Intro Calculus) LESSON 2: LIMITS
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LESSON 2: LIMITS (Intro Calculus)
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(Intro Calculus) LESSON 2: LIMITS
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LESSON 2: LIMITS (Intro Calculus)
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(Intro Calculus) LESSON 2: LIMITS
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LESSON 2: LIMITS (Intro Calculus)
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(Intro Calculus) LESSON 2: LIMITS
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LESSON 2: LIMITS (Intro Calculus)
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(Intro Calculus) LESSON 2: LIMITS
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LESSON 2: LIMITS (Intro Calculus)
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(Intro Calculus) LESSON 2: LIMITS
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(Intro Calculus) LESSON 2: LIMITS
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(Intro Calculus) LESSON 2: LIMITS
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(Intro Calculus) LESSON 2: LIMITS
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LESSON 2: LIMITS (Intro Calculus)
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(Intro Calculus) LESSON 2: LIMITS
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LESSON 2: LIMITS (Intro Calculus)
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17.(a)
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67
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17.(b)
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68
LIMITS PRACTISE PROBLEMS (Intro Calculus)
Homework:
 Study the lesson thoroughly until you can do all of questions 1 to 16 on page 31
from start to finish without any assistance.
 Do all of the Practise Problems below (solutions are on pages 70 to 78).
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Practise Problems:
For each of the following questions, find the value of the limit, if it exists. If it does not exist, is
it infinity, negative infinity, or neither? Justify your answers.
4.
lim
x 2  5x  6
x3
lim
x2  4
x2
x  3
x  2
x 2
5. lim
x 1
x2  4
7. lim 2
x  2 x  x 6
x3
lim
x 2
x 2  3x  2
x 1
11.
lim
x  2
x2  4
x 2
 2x  6
14. lim
x  2
x 3  3x 2
16. lim
2
x  3 x 9
19. lim
x 3
22.
lim
x 2  2x  3
x2  x  6
x   8
24. lim
x 3
x  1
x 7
x  6x  7
x 7
23.
lim
x  3
2
5 x  15
x  2 x  15
2
 x2
2x 
lim 

x  2  x  2
x  2 
lim
x3  8
x2
x 3
x 9
x  2
25. lim
x 2
x 2
26. lim
28. lim
x 1 1
x
29. lim
x 5 3
x 4
25  x 2  3
4x
32. lim
2x  3  3x
x 3
x2
3  5x  1
x 2
31. lim
33. lim
x 5
3x  1  4
34. lim
© 1997-2011 Grant Skene for
15. lim
21.
lim
30. lim
x 5
x2  x  2
2x 2  3x  2
x 2  2x  1
x2  1
20.
lim
x  1
x3  x2
x2  1
12. lim
18.
x
1  3x  1
x2
9.
x2  4 x  5
17. lim
x 5
x 5
27. lim
x 0
x  x 6
14 x  13 
 x 1
 x  8  x 2  5 x  24 
x2  9
x 3  27
x6
5
2
x 3
x 2  4 x  12
x 6
6. lim
x2
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13.
2 x
x2  4
x2  4
x 2
3. lim
x 3  2x 2
8. lim
x 2
x 2
x 3  2x 2  3x
x2  9
10. lim
x2  1
x 1
2. lim
PL
1.
x0
x 4
x 0
3x
2 4  x
x9
x4
x3
35. lim
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x 0
x
x
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(Intro Calculus) LIMITS PRACTISE PROBLEMS
lim
x 1
x2  1
42. lim
x 2  16
2 x
39.
x  1
x 4
45.
48.
51.
x  3
lim
x  x  2
2 x
lim
x  2
54. lim
x 2  12 x  36
x 6
x   2
60.
2x 3  7 x
3x 3  4 x 2  1
lim
x  
3x 3  1
2x 2  x 4
x
x 
44.
46. lim
x2  4 x  3
x2  6x  9
47.
50.
52.
55.
lim
x 2  5x  4
53.
lim
2x 3  2
56.
x  4
x  1
58. lim
x
61.
64.
66.
9x2  2
3  4x
67.
69. lim
x 
71. lim
x 
73.
x
2
3x  2x 2  1
1  2x
h0
lim
x 2  2x  3
x4  x2  6
x  
lim
x  
7 x
x2  6x  5
lim
x  x  5
x   5
2
lim
x 5
lim
x 2  2x  3
lim
1
x 1
x  5
x  
2
5
 1  1
5
 1 x 2  2
x  100
4
3h  2sin h
h
© 1997-2011 Grant Skene for
74. lim

77. lim
tan  3 x 
x
x
x0
x 2  3x  1  x
x  1
1  x2
1  x2
lim
x  
2 x 6  5 x 3  10
6x  4 x3
x2  2
x 7
x 
 2x
68. lim
 2x
70. lim
x
x 
72.

x  
x 0
78.
Grant’s Tutoring (www.grantstutoring.com)
 1  x  1
4
2
 3 4  x 2 
2
 x 3  x 2  x  5
 x  1
lim
75. lim
3
100  x 4
x 
 2 x 2  1
 3  x
x  3
65. lim
5x 2  4
 3x
lim
62.
x2  5
x 2  2x  4
2
x  3
x
1  x 
2
x2  4
x 2  5x  6
lim
59. lim
 7  x 2  5 x  2
 2x
lim
x  
x
76. lim
2
x 2  16
2 x
10
x


49. lim  2

x  3  x  x 6
 2x 
 3x
x  1
x 4
x 3
2
x  2x  3
2
x  
1 x
1 x
lim
41. lim
43. lim
63. lim
lim
38.
4 x 2  5x  6
SA
M
x
3
x4
x4
x 2  9 x  22
lim
x 3
x 2  5x  6
x   2
x  4
x 3
x2  4 x  3
x2  6x  9
lim
57. lim
40.
lim
PL
x 1
37.
E
x 1
x 1
36. lim
69
3x 2  5
x2  5
sin x  3 x
2x
lim
x  1
2
4 x2  1
4 x 2  8x  3
DO NOT RECOPY
70
0
x 2  5x  6
is a
limit with “ x  3 ” as the factor on top
x  3
x3
0
and bottom:
lim
x  3
2.
3.
4.
lim
x2  1
5

x 1
3
lim
x2  4 0
  0
x2 4
x 2
x 2
 x  3   x  2
x3
 1
lim
x 2
13.
x  2
lim
 x  2  x  2
x2  4
 lim
x  2
x2
 4
x2
lim
x3
0
x  3x  2
is a
limit with “ x  1 ” as the factor on top
x 1
0
and bottom:
x 1
 x  6   x  2
x 2  4 x  12
 lim
x 6
x 6
x 6
x 6
x2  4
x2  x  6
x 2
8.
3
 lim
x 2
 x  2  x  2

 x  2  x  3
lim
x 7
x 2  x  2
x 3  2x 2
lim
 lim
 4
x 2
x  2  x  2
x 2
x  1
2
x 1
is a
x3  x2
0
limit with “ x  1 ” as the factor on top
0
x  1
x2  1
x 2  x  1
 lim
x  1
 x  1  x  1

18.
1
1
 
2
2
x 3  2x 2  3x
0
10. lim
is a
limit with “ x  3 ” as the factor on
x 3
0
x2  9
top and bottom. Don’t miss the common factor of “x” that you
can pull out of the top:
lim
x3
 lim
x3
11.
lim
x  2
x 3  2x 2  3x
2
x 9
 lim
x  x  3   x  1
 x  3  x  3
x3


x x 2  2x  3
x2  4
8

 2
x 2
4
© 1997-2011 Grant Skene for
x 3  3x 2
x2  9
x 3

1
4
4
2  x  3 
 x  3 x  2
4

32  0 
 0
5
0
limit with “ x  7 ” as the factor on top
0
is a
 lim
x 7
x 7
 x  7   x  1

1
8
0
limit with “ x  3 ” as the factor on top
0
is a
and bottom. Don’t miss the common factor of “ x 2 ” that you
can pull out of the top:
lim
x 3  3x 2
2
x 9
x 3
 lim
x 2  x  3
x 3
 x  3  x  3

9
3

6
2
0
x2  4 x  5
is a
limit with “ x  5 ” as the factor on top
x 5
0
and bottom:
x 5
lim
x 5

20.
x2  4 x  5
 lim
x 5
x 5
 x  5  x  1
x 5
 6
0
5 x  15
is a
limit with “ x  3 ” as the factor on
0
x 2  2 x  15
top and bottom. Don’t miss the common factor of “5” that you
can pull out of the top:
lim
x  3
5 x  15
lim
x  3
19. lim
x2  9
34 
 2
6
x  2
17. lim
and bottom. Don’t miss the common factor of “ x 2 ” that you
can pull out of the top:
lim
x 7
x2  6x  7
16. lim
4
5
x3
5
 x  3  x  2
x2  6x  7
and bottom:
2
x3  x2
  x  2

 x  2  x  2
 lim
5
 lim
x2  x  6
x 7
15. lim
and bottom. Don’t miss the common factor of “ x ” that you
can pull out of the top:
lim
x2  4
x  2
x 7
 8
2
9.
x  2
 lim
25  x  3
x3
0
x  2x
is a
limit with “ x  2 ” as the factor on top
x 2
0
lim
x 2
0
limit with “ x  2 ” as the factor on top and
0
is a
 2 x  6 5
 lim
SA
M
lim
3
5
 2 x  6 5
x3
0
is a
limit with “ x  2 ” as the factor on top
7. lim 2
x 2 x  x 6
0
and bottom:
x2  4
2 x
x2  4
lim
0
x 2  4 x  12
is a
limit with “ x  6 ” as the factor on top
x 6
x 6
0
and bottom:
lim
lim
x 2
0
is a
limit with “ x  3 ” as the factor on top
0
x2  x  6
and bottom. Don’t miss the common factor of “2” that you can
pull out of the top:
14. lim
 x  1  x  2
x 2  3x  2
lim
 lim
 1
x 1
x 1
x 1
x 1
6.
x2  4
x  2
2
lim
2 x
lim
x  2
 x  2  x  1

 x  2  2 x  1
 lim
2x 2  3x  2
x  2 ” distract you. That is only relevant if you need to
establish the sign of zero in a problem:
0
x 4
is a
limit with “ x  2 ” as the factor on top and
x2
0
bottom:
lim
x2  x  2
bottom. Do not let the fact that you are given a one-sided limit “
2
x  2
5.
x 2  5x  6
 lim
x  3
x3
0
limit with “ x  2 ” as the factor on top
0
is a
2x 2  3x  2
and bottom:
x 2
E
lim
x2  x  2
12. lim
PL
1.
SOLUTIONS TO LIMITS PRACTISE PROBLEMS (Intro Calculus)
2
x  2 x  15
x 2  2x  3
x2  x  6
and bottom:
x3
lim
x 3
lim
x  1
x 2  2x  3
x2  x  6
Grant’s Tutoring (text or call (204) 489-2884)
x  3
is a
x 3

5  x  3
 x  3  x  5

5
5
 
8
8
0
limit with “ x  3 ” as the factor on top
0
 lim
x 2  2x  1
x2  1
 lim
 x  3  x  1

 x  3  x  2
4
5
0
 0
2
DO NOT RECOPY
(Intro Calculus) SOLUTIONS TO LIMITS PRACTISE PROBLEMS
21.
 x2
k k
2x 
lim 

 is “  ”. Pull the two fractions
0 0
x  2  x  2
x  2 

0
x 2
is a
limit with “ x  2 ” as the factor on top
0
x 2
and bottom. To be able to factor the top, however, we must
remove the square roots by multiplying top and bottom by the
25. lim
x2
together over a common denominator (which is clearly “ x  2 ”)
and reassess the problem:
 x2
 x 2  2x 
0
2x 
limit
lim 

  lim 
 which is a
x  2  x  2
x


2
0
x  2 

 x  2 
conjugate “ x  2 ”:
x  x  2
x  2x
 lim
 2
x  2
x2
x2
k k
14 x  13 
 x 1
x 8  2
 is “ 0  0 ”. Pull the two
x  5 x  24 

lim
x   8


lim
x   8
lim
x   8
 x 1 x  3
14 x  13 




 x  8 x  3  x  8  x  3 
  x  1 x  3
14 x  13 



  x  8  x  3  x  8  x  3 
  x  1 x  3  14 x  13 


 x  8  x  3


0
x 3
is a
limit with “ x  9 ” as the factor on top and
0
x 9
bottom. To be able to factor the top, however, we must remove
the square root by multiplying top and bottom by the conjugate
x9
“ x  3 ”:

lim
x   8
23.
lim



is a

a3  b3   a  b a2  ab  b2
lim
x 3
x2  9
3
x  27
 lim
x 3

 x  3  x  3
6


 x  3  x 2  3 x  9 27
3x

1 1
2

3
3
“ x  1  1 ”:
lim
x0
2
9
 lim
x0
29.
x 1 1

x
x

x 1 1
x11
 lim
x 1 1 x  0 x x 1 1

x
x 1 1



1
1

2
1 1
0
x 53
is a
limit with “ x  4 ” as the factor on top
0
x 4
and bottom. To be able to factor the top, however, we must
remove the square root by multiplying top and bottom by the
lim
x 4
conjugate “ x  5  3 ”:
lim
x4
 lim
x 53

x 4
x 53
x 59
 lim
x  5  3 x  4  x  4 x  5  3

x 4
x4
 x  4 
x 53



1
1

6
93
0
3  5x  1
is a
limit with “ x  2 ” as the factor on top
0
x 2
and bottom. To be able to factor the top, however, we must
remove the square root by multiplying top and bottom by the
30. lim
x 2
conjugate “ 3  5x  1 ”:
lim
x2
 lim
x2
© 1997-2011 Grant Skene for
1  3x  1

0
x 1 1
is a
limit with “x” as the factor on top and
0
x
bottom. To be able to factor the top, however, we must remove
the square root by multiplying top and bottom by the conjugate
  x  8   x  2 
6
6



11
  x  8  x  3  11


 x  2 x 2  2 x  4
x3  8
 lim
 12
x  2 x  2
x2
x2  9

28. lim
0
limit with “ x  3 ” as the factor on top and
x  3 x 3  27
0
bottom. Recall how to factor a Difference of Cubes:
24. lim
1
1

6
93
x 0
a3  b3   a  b a2  ab  b2


x0
 x 2  10 x  16 
0
limit with “ x  8 ” as

 which is a
0
  x  8 x  3 
x   8


x 1  3x  1
x
1  3x  1

 lim
1  3x  1
1  3x  1 x  0 1  3x  1
x
 lim
0
x3  8
is a
limit with “ x  2 ” as the factor on top and
x  2 x  2
0
bottom. Recall how to factor a Sum of Cubes:
x  2
x 3
0
x
is a
limit with “x” as the factor on top and
0
1  3x  1
bottom. To be able to factor the bottom, however, we must
remove the square root by multiplying top and bottom by the
lim
lim
lim
x 9
 x  9 
x 0
x0
 x 2  4 x  3  14 x  13 


 x  8  x  3


 x 2  10 x  16 


  x  8 x  3 
lim
x 3
 lim
x 3 x 9
27. lim
the factor on top and bottom:
x   8
x 3

x 9
lim
x 9
conjugate “ 1  3x  1 ”:
SA
M

lim
x   8

1
1

2 2
2 2

26. lim
Multiply the top and bottom of the first fraction by “ x  3 ” to
create the lowest common denominator of “  x  8 x  3 ”.
x   8

x 2
PL
fractions together over a common denominator and reassess the
problem. Note that “–8” is a zero of both denominators so, “
x  8 ” is a factor of both denominators. Factor to more clearly
see the common denominator:
 x 1
14 x  13 
14 x  13 
 x 1
lim 
 2


  lim 
x  5 x  24  x   8  x  8  x  8  x  3 
x   8  x  8
lim
 x  2
x2
E
22.
x 2
x 2
x 2
 lim
2
lim
x 2

x 2
lim
x2
with “ x  2 ” as the factor on top and bottom. Don’t miss the
common factor of “x” that you can pull out of the top:
x  2
71
9   5 x  1
3  5x  1 3  5x  1

 lim
x 2
3  5 x  1 x  2  x  2 3  5 x  1

5 x  10
 x  2  3 
Grant’s Tutoring (www.grantstutoring.com)
5x  1

 lim
x2
5  x  2
 x  2  3 
5x  1


 
5
6
DO NOT RECOPY
72
25  x 2  3
0
is a
limit with “ x  4 ” or, in this case, “
x 4
4x
0
4  x ” as the factor on top and bottom. To be able to factor the
top, however, we must remove the square root by multiplying
lim
x0
We must, however, get rid of the absolute value signs before we
can cancel the factors.
 u if u is positive
Recall: u  
u if u is negative
top and bottom by the conjugate “ 25  x 2  3 ”:
lim
x4
 lim
x4
 lim
x4
 lim
x4
25  x 2  3
25  x 2  3

4x
25  x 2  3
Use one-sided limits to establish the sign of zero:
x  0
2
25  x  9
4  x 

25  x 2  3
16  x 2
4  x 

25  x 2  3

4  x 

25  x  3
lim

Therefore, lim
x0
conjugate “ 2x  3  3x ”:
 lim
x3
 lim
x3
 lim
36. lim
x 1
2x  3  3x
x  3
 x  3 
2x  3  3x

 x  3 
2x  3  3x
x  1
0


0
x 5
is a
limit with “ x  5 ” as the factor on top
0
3x  1  4
and bottom. To be able to factor the bottom, however, we must
remove the square root by multiplying top and bottom by the
conjugate “ 3x  1  4 ”:
 lim

 x  5 
3x  1  4
3 x  15
x  1

lim
x 5
 x  5  3 x  1  4 

3  x  5

37.
8
3
0
3x
is a
limit with “x” as the factor on top and
0
2 4  x
bottom. To be able to factor the bottom, however, we must
remove the square root by multiplying top and bottom by the
x 0
conjugate “ 2  4  x ”:
x0
 lim
x0

3x 2  4  x
3x
2 4  x

 lim
2  4  x 2  4  x x  0 4  4  x 

3x 2 4  x
x
x 1
 x  1
1
(since x  1  positive as x  1 )
Therefore, lim
34. lim
lim
2
x 1
x 1
does not exist because the
x 1
Left-hand Limit  the Right-hand Limit (LHL≠RHL)  1  1 .
 x  5 3 x  1  4
x 5
3x  1  4

 lim
3 x  1  16
3x  1  4
3x  1  4 x  5
x 5
x 1
 lim
x  1 x  1
lim
x 5
lim
 1 
(since x  1  negative as x  1 )
1
1
 
6
9 9
33. lim
x 5
x  1
x 1
x 1
 lim
 1
x  1 x  1   x  1
lim
x  1

  x  3
0
x 1
is a
limit with “ x  1 ” as the factor on top and
0
x 1
bottom. Get rid of the absolute value signs. Use one-sided
limits to establish the sign of zero:
SA
M
x3
Left-hand Limit  the Right-hand Limit  1  1 .
2x  3  3x
2x  3  3x

x 3
2x  3  3x
2x  3  3x
 x  3 
x
does not exist because the
x
PL
0
2x  3  3x
is a
limit with “ x  3 ” as the factor on
0
x 3
top and bottom. To be able to factor the top, however, we must
remove the square roots by multiplying top and bottom by the
x 3
lim
x
x
 lim
 1 (since x  positive as x  0 )
x x  0 x
x  0
8
8
4
 
3
93 6

1
x
x
 lim
 1 (since x  negative as x  0 )
x x  0  x
x  0

4  x  4  x 
2
lim
x  0
 0 
–1
32. lim
x3
0
x
is a
limit with “x” as the factor on top and bottom.
0
x
35. lim
E
31.
SOLUTIONS TO LIMITS PRACTISE PROBLEMS (Intro Calculus)
  32  4  
1

38.
lim
x  4
x4
0
is a
limit with “ x  4 ” as the factor on top
x4
0
and bottom. Get rid of the absolute value signs. Use the given
one-sided limit to establish the sign of zero:
x   4
 –4
–5
lim
x   4
x4

x4
lim
x   4
  x  4
x4
–3
 1
(since x  4  negative as x   4  )
lim
x  1
0
1 x
is a
limit with “ x  1 ”, or “ 1  x ”, as the factor
0
1 x
on top and bottom. Get rid of the absolute value signs. Use the
given one-sided limit to establish the sign of zero:
x  1
 12
0
lim
x  1
1 
2
1 x
1 x
 lim
 1
1  x x  1  1  x 
(since 1  x  negative as x  1 )
© 1997-2011 Grant Skene for
Grant’s Tutoring (text or call (204) 489-2884)
DO NOT RECOPY
(Intro Calculus) SOLUTIONS TO LIMITS PRACTISE PROBLEMS
39.
x 1
lim
2
x  1
x 1
is a
0
limit with “ x  1 ” as the factor on top and
0
42.
bottom. Factor the bottom first and foremost:
x 1
x 1
.
lim
 lim
2
x  1 x  1
x  1  x  1 x  1
k
x 2  16
is a
limit (the answer is either  or –). Use
x 4 2 x
0
one-sided limits to establish the sign of zero:
lim
x  4
again we see that the bottom is  0 2 . Clearly, the second
lim
x  4
factor is positive (since it is “2”). Use the one-sided limit to
establish the sign of zero in the first factor:
 1
lim
x 2  9 x  22
lim
4 x 2  5x  6
x   2
0
limit with “ x  2 ” as the factor on
0
is a
top and bottom. Factor first and foremost:
x 2  9 x  22
lim
x   2
2
4 x  5x  6

lim
x   2
 x  2 x  11
 x  2 4 x  3
x4
43. lim
x 3
x3
x 2  2x  3
SA
M
 x  2 x  11
x  2  x  2 4 x  3
(since “ x  3 ” is negative as x  3 )
lim
x  3
  positive  negative   negative
 lim
x  2
 x  2  4 x  3

  13
13
 
11
11
2
0
x  16
41. lim
is a
limit with “ x  4 ” as the factor on top and
x 4 2 x
0
bottom. Use the conjugate “ 2  x ” to get rid of the square
root on the bottom:
 x  4  x  4   2  x
x 2  16
lim
 lim
x 4 2 x
x4
2 x
2 x
 lim
x4
 lim
x4
 x  4  x  4   2 
x
Therefore, lim
x 3
44.
lim
x
   8 4  
1
 32
x  3
4
6
x  5x  6
is a

x3
x 2  2x  3

lim
x  3
 
 positive
 0    4        
x3
x2  4
lim
x  3
x3
x 2  2x  3

does not exist (LHL  RHL).
k
limit (the answer is either  or –).
0
First and foremost, we know that “ x  3 ” is a factor of the
bottom since “3” is a zero of the bottom. Factor and reassess the
limit (we have no need to factor the top), using the one-sided
limit to establish the sign of zero:
x  3
2
lim
x  3
 3
4
x2  4
5
x2  4
is

 lim
2
 0   1   
x  5 x  6 x  3  x  3 x  2
(since “ x  3 ” is negative as x  3 )
lim
x  3
© 1997-2011 Grant Skene for
is
x 2  2x  3
2

4x
 x  4   x  4 2 
  x  4
x3
 x  3 x  1
(since “ x  3 ” is positive as x  3 )
   x  2 x  11
  x  2  x  11
x  3
 3 
2
 positive   13 negative 
lim
k
limit (the answer is either  or –).
0
x3
6

is
lim

 negative
 0    4        
x  3  x  3 x  1
x  2 is positive as x   2 ; therefore:
 x  2 x  11  negative
is a
x  3
–1
–2 
x 2  16
 
2 x
Use one-sided limits to establish the sign of zero:
x   2
Thus,
x  4
First and foremost, we know that “ x  3 ” is a factor of the
bottom since “3” is a zero of the bottom. Factor and reassess the
limit:
x3
x3
6 
6
is
lim
 lim
x  3 x 2  2x  3
x  3  x  3 x  1
 0 4    0  ? 4   
factor is negative (since it is “–13”). Use the one-sided limit to
establish the sign of zero in the first factor:
 x  2  x  11
lim
x 2  16
does not exist (LHL  RHL;   –).
2 x
Therefore, lim
Establish the sign inside the absolute value. Subbing in x = –2
again we see that the top is  0 13 . Clearly, the second
–3
x 2  16

2 x
32  32  positive
x  16
is
 negative
0  negative
2 x
PL
   x  1 x  1
x 1
x 1
1
1
 lim

 
2
 x  1 x  1 x  1   x  1  x  1 2
x  1
40.
  negative  positive   negative
 x  1 x  1  negative
x  4
(since “ 2  x ” is negative as x  4  )
 negative   2 positive 
Thus,
lim
2
x  4
x  1 is negative as x  1 ; therefore:
 x  1  x  1
(since “ 2  x ” is positive as x  4  )
lim
2
5
32  32  positive
x 2  16
is
 positive
0  positive
2 x
E
x  1
x  4
 4 
3
Establish the sign inside the absolute value. Subbing in x = 1
0
73
x2  4
2
x  5x  6
Grant’s Tutoring (www.grantstutoring.com)
 lim
x  3
x2  4
 
 x  3 x  2
DO NOT RECOPY
74
x2  4 x  3
lim
2
x  6x  9
x  3
0
limit with “ x  3 ” as the factor on top
0
is a
10
x


49. lim  2

x 3  x  x 6
and bottom.
10
2
x  4x  3
lim
 x  3   x  1
 x  3  x  3
 lim
x2  6x  9
x  3
x  3
10
x


lim  2

 x  x 6

x 3
lim
lim
50.
4
x2  4 x  3
Use one-sided limits to establish the sign of zero:
x 1
 
x 3
 lim
x  3
x2  4 x  3
x  6x  9
6  
 0 2  
lim
x 3
47.
 lim
2
x 3
–6
 lim
x2  6x  9
lim
x   5
x  x  5
x2  4 x  3
 x  3
x 3
x2  6x  5
2
is a
2
Therefore,
x  x  5
2

lim
51.
lim
x  2
x 1
x  x  5
x  x  5
2

lim
x   5
lim
x  2
52.
–6
–5 
–4
x 1
4  
is
  (since “ x  5 ” is
x  x  5
 5    0   
48.
lim
x   2
x2  6x  5
2
x  x  5

x 2  5x  6
x  x  2
3
x 1
 
x  x  5
lim
x   5
0
is a
limit with “ x  2 ” as the factor on
0
top and bottom.
x 2  5x  6
lim

3
x   2 x  x  2
which is now a
lim
x   2
 x  2  x  3
x  x  2
3
2

2 x  0
x3
2
1
lim
is
  (since “  x  2 ” is
2
x   2 x  x  2
 2    02  


clearly positive, being squared)
x 2  5x  6
lim

3

x  x  2
x   2
x 3
lim
 
2
x   2 x  x  2
© 1997-2011 Grant Skene for
53.
3
 as x  2 )
(since it is
x 2  5 x  4 is a “ 0 ” limit. First and foremost,
lim
x  4
factor since we know that “ x  4 ” is a factor. Then use the onesided limit to establish the sign of zero:
x   4
 –4
–5
2
x  5x  4 
lim
x   4
lim
x   4
–3
 x  4  x  1
“ x  4 ” is negative as x   4  , so
 0   3          
x 2  5x  4  0
lim
x  4
lim
x  3
 x  4  x  1
lim
x  4
is
 ; the limit exists.
(since it is
)
x 2  2 x  3 is a “ 0 ” limit. First and foremost, factor
since we know that “ x  3 ” is a factor. Then use the one-sided
limit to establish the sign of zero:
x  3
x3
lim

x
x   2
 x  2 2
k
limit (the answer is either  or –).
0
 2
1
positive as x   5 )
x   5
x  5 does not exist
x  2
x   5
lim
lim
x  5
establish the sign of zero:
k
limit (the answer is either  or –). Use the
0
one-sided limit to establish the sign of zero:
x   5
)
2  x is a “ 0 ” limit. Use the one-sided limit to
which is now a
lim
–4
 –5 
 
 x  5  x  1
x  5
x   5
because it cannot be approached from both sides.
SA
M
lim
x  5

x  5 does not exist (since it is
x   5
0
limit with “ x  5 ” as the factor on
0
top and bottom.
x2  6x  5
lim
x2  4 x  3
x2  4 x  3
is
 lim
 x  3 x  3 x  3  x  32
  (since the bottom is squared and so positive)
x2  4 x  3

PL
x 3
x2  4 x  3

0
if

0

does not exist if
x   5
x 3
lim
x  5 is a “ 0 ” limit.
Recall:
k
is a
limit (the answer is either  or –)
0
x2  6x  9
with “ x  3 ” as the key factor on the bottom (we have no need
to factor the top). Factor and reassess the limit:
46. lim
lim
x  5
 
2
x 1
is
  (since “ x  3 ” is negative as x  3 )
0
x 3
x2  6x  9
x  3
x 3
 3
2
10
x


3
lim 
 is   which is clearly positive because it
x  3  x2  x  6 
0
is raised to the power of 10. All even powers produce positive
answers.
k
which is now a
0
limit (the answer is either  or –). Use the one-sided limit to
establish the sign of zero:
x  3
k
limit (the answer is either  or –)
0
is a
E
45.
SOLUTIONS TO LIMITS PRACTISE PROBLEMS (Intro Calculus)
2
lim
x  3
 3
2
x  2 x  3  lim
x  3
4
 x  3 x  1
“ x  3 ” is negative as x  3 , so
 0   4          
lim
x  3
lim
x  3
 x  3 x  1
 ; the limit does not exist.
x 2  2 x  3 does not exist (since it is
Grant’s Tutoring (text or call (204) 489-2884)
is
)
DO NOT RECOPY
(Intro Calculus) SOLUTIONS TO LIMITS PRACTISE PROBLEMS
x 2  12 x  36 is a “ 0 ” limit. First and foremost,
54. lim
x 6
58.
factor since we know that “ x  6 ” is a factor.
x 2  12x  36  lim
x 6
 02
which is clearly
 x  62
lim
x 6
  since anything squared will be
0  0 if the limit is
positive. We know
.
x 2  12 x  36  0
Therefore, lim
x 6
55.
 x  6 x  6 
2 x 3  2 is a “ 0 ” limit. First and foremost, factor
lim
x  1
since we know that “ x  1 ” is a factor. Don’t miss the common
factor of “2” that we can pull out:
lim
x 1
2 x 3  2  lim
x 1


2 x 3  1  lim
x 1


59.

2  x  1 x 2  x  1

3
3
2
2
(Recall: a  b   a  b a  ab  b )
is
x 1
 2    0    3        
We know
Therefore,
56.
lim
lim
x  1
0  0 if the limit is
lim
x  1


60.

.
x  
lim
3x 3  4 x 2  1
and bottom:
x
lim
x
 lim
61.
3x 3  4 x 2  1
x
x  
 3x 3 1 
x3  3  3 
 x
x 

2
 2x
x4 
x4  4  4 
 x
x 

1
x3
x 2  2x  3
lim
x  
x4  x2  6

30
 0
  0  1
(since
3
 0)

is solved by factoring “ x 2 ” out of the top
and “ x 4 ” out of the bottom:
 ) it will always
2
x  2x  3
lim
x  
x4  x2  6
 lim
x  
 x 2 2x
3 
x2  2  2  2 
x
x
x 

4
2
2x
x
6 
x4  4  4  4 
x

x
x


2 3

100
x x2

 0
1
6   1  0  0 
2
x 1  2  4 

x
x 
1
 lim
x  
3
 lim
 lim
 2

x  2  1
x

x  
is solved by factoring “ x ” out of the top
2x  7 x
2
3x  1
2x 2  x 4
3
1
 
x 1
3
is solved by factoring “ x 3 ” out of the top and
3
.
give you a positive answer ( 4 is positive 2; 9 is positive 3;
and so on).
1
1
  since a square root will be positive.
is
lim
0
x 1
x  1
57.
x
2x 2  x 4
lim
SA
M
(keeping in mind that you can’t compute
2x  7 x
 1
x2 
1
x2  2  2 
1
x
2
x 
0 1

x
 lim

 1
x2  x   1  1 0  1
2  1
x  2  2
2
x
x
x 

 lim
3x 3  1
x  
answer is either  or –).
Note, the answer for a square root is always positive. If you ask
your calculator to compute the square root of any number
3
1  x2
lim
0  0 if the limit is  .
k
1
is now a
limit (the
But that now means lim
0
x 1
x  1
We know
x  1
2
“ x 4 ” out of the bottom:
2x 3  2  0
“ x  1 ” is positive as x  1 , so we have
lim
1 x
x
1
is a “ 0 ” limit.
x 1
Therefore,
is solved by factoring “ x 2 ” out of the top and
1  x2
bottom:
lim
2  x  1 x 2  x  1
1  x2
lim
x
PL
“ x  1 ” is positive as x   1 , so
3x  2x 2  1
is solved by factoring “ x 2 ” out of the top
x 
1  2x
and “x” out of the bottom:
 3x 2x 2
1 
x2  2  2  2
x
2
x
x 
3x  2x  1

lim
 lim
x
x
1  2x
 1 2x 
x 

x 
x
1 
3
x  2 2 
x

x
    0  2  0   
 lim
1
x
2
2
x
lim
E
lim
x 6
75
62.
 2x 3 7 x 
x3  3  3 
 x
x 

3
2

3x
4x
1 
x3  3  3  3 
 x
x
x 

2 x 6  5 x 3  10
lim
6x  4 x3
x  
is solved by factoring “ x 6 ” out of the
top and “ x 3 ” out of the bottom:
lim
6
2 x  5 x  10
6x  4 x3
x  
7
20
2
x
 lim


4 1
x
3

0

0
3
3  3
x x
3
2
x  
lim
x 
x
x  
5 10 

x3  2  3  6 

x
x     2  0  0     
6
04
4
4
x2
 lim
63.
 lim
3  2x6
5 x 3 10 
x6  6  6  6 
 x
x
x 

3
 6x 4 x 
x3  3  3 
x
x 

2

2
 2 x is solved by factoring out “ x ”:
 x 2 2x 
2

lim x 2  2 x  lim x 2  2  2   lim x 2  1     1  0  
x
x 
x
x  x  

x 
© 1997-2011 Grant Skene for


Grant’s Tutoring (www.grantstutoring.com)
DO NOT RECOPY
64.
SOLUTIONS TO LIMITS PRACTISE PROBLEMS (Intro Calculus)
7 x
lim
is solved by factoring “ x 2 ” out of the
2
x  
5x  4
68.
bottom:
7 x
lim
2
x  
5x  4
x  
But:
65.
2
x  5
 5x
4 
x2  2  2 
 x
x 

2
x  
7 x
 lim
4
x  
x2
x2  5 
 x 5
4

7

 50
2 
x  2  2
x

x


 lim
x
7
x


x  
x x
x
2
x  1
x
9x2  2
 lim
x  
3  4x
But:
x2  9 
x2  9 
2
x2
3

x  4
x

x  

2
x 9
2


is solved by factoring the highest
x 4  100
power of “x” out of each piece:
 2  3x 2
1   2  x 2
2 
 x  2  2    x  2  2  
3x 2  1 x 2  2
x     x
x  
  x
lim
 lim 
4
x  
x  


x 4  100
x
100
x4  4  4 
x
x 

lim
x  



 lim
1 
2 


x2  3  2   x2 1  2 

x 

x  but, on top, x 2  x 2  x 4 :
100 

x 4 1  4 

x 
 lim
1 
2 

x  3  2  1  2 

x 
x    3  0 1  0   3
100 
1 0
4 
x 1  4 

x 
x  
4
x  


 2 x2  1




is solved by factoring the highest

© 1997-2011 Grant Skene for


2 
1 

x 4  3  2  1  2 

x 
x    3  0 1  0   3
7 
5 2 
1  0 1  0  0 

x 4  1  2  1   2  
x x 

x 
 lim
x
x  x   x since x   
3x 2  1 x 2  2
2
x 2  7 x 2  5x  2
2
x 2  lim
x2   9  0  3
x  
04
4
3

3

x  4
x   4
x

x

lim

 3x
2 
1 


x2  3  2   x2 1  2 

x


x
 but x 2  x 2  x 4 :
 lim
7  2
5 2 
x  2
x 1  2   x 1   2 
x x 

x 

SA
M
lim
 9x2
2 
x2  2  2 
 x

x


 lim
x  
 3 4x 
x 

x 
x
x

9x2  2
is solved by factoring “ x 2 ” out of the top
3  4x
and “x” out of the bottom:
lim
x  
x  
lim
power of “x” out of each piece:
 2  3x 2
2   2  x 2
1 
 x  2  2    x  2  2  
2
2
3x  2 x  1
x
x
x
x
 
   
 
lim
 lim 
2
x  x 2  7 x 2  5 x  2
x  
7   2  x 2 5x
2 
2 x
 x  2  2    x  2  2  2  
x     x
x
x  
  x
2
2
x 1 2
x 2  lim
x  10  1
7
7
x
10


x 1  
x 1  
x
x


x2  1 
lim
x
69.
x 2  x  x since x  
But:
 lim
2
x2
7


x 1  
x

1 
1

x 4  2  3  1  
x   2  0 1  0 

x 

 2
0 1

4  100
x  4  1
 x

PL
x2  2
 lim
x
x 7
x

1  
1

x 3  2  3   x 1  
x


x

 lim
but, on top, x 3  x  x 4 :
x
 100

x 4  4  1
 x

7
5
x2  2
is solved by factoring “ x 2 ” out of the top and
x  x 7
“x” out of the bottom:
lim
67.
is solved by factoring the highest power
100  x 4
of “x” out of each piece:
 3  2x 3
1    x 1 
 x  3  3    x    
2 x 3  1  x  1
x     x x  
  x
lim
 lim 
4
x
x
 100 x 4 
100  x
x4  4  4 
 x
x 


4
x2
x2
2

 1  x  1
3
x
lim
2
66.
7 x
 lim
x 2  x   x since x   
7 x
lim
x  
7 x
 lim
 2x
lim
E
76
70.
lim
x
 2x
x
4
2

 x
 3 4  x2
 x3
2

2
 x 5

is solved by factoring the highest
power of “x” out of each piece:
lim
x
2x
x
4
2

 x
 3 4  x2
 x3
2

2
 x 5

2
 2  2x 2
3   2  4
x 2 
 x  2  2    x  2  2  
x     x
x  
  x
 lim 
4
3  
2
x 


x
x
5 
4 x
2 x
 x  4  4    x  2  2  2  
x     x
x
x  
  x
2
3 

 4

x 2  2  2   x 4  2  1

x


x

 lim
1
1 5 
x 4 

x 1    x 2 1   2 
x
x x 


2
3  4


x6  2  2   2  1
2

x  x
   2  0  0  1  2
 lim
1 
1 5 
x 6 
1  0 1  0  0 
x 1   1   2  
x 
x x 

Grant’s Tutoring (text or call (204) 489-2884)
DO NOT RECOPY
(Intro Calculus) SOLUTIONS TO LIMITS PRACTISE PROBLEMS
71.
lim
1  x 
x2  5
2
x
x  2x  4
“x” out of each piece:
is solved by factoring the highest power of
73.
lim
x  
 2x
x
x
72.
lim
 x  1
3x 2  5
2
x 5
of “x” out of each piece:
x  
lim
 x  1
x  
2
3x  5
x2  5
 lim
x  
x  
 2x
x
2
 x 
 1  1
3
5
2
 2  2x 2
3 
 x  2  2    x
x  
  x
5
lim
x  
5
 2  x2
1 
 x  2  2    1
x  
  x
5
3 

x10  2  2   x

x 
We are still not ready to cancel! We
5
1 
10 
x 1  2   1

x 
E
x  
 lim
x  
  x 1  2  3 x 2
5 
 x  x  x   x  2  2 

x
x
 


2
5 
2 x
x  2  2
x
x 

x 2  x   x since x  
SA
M
x  
lim
 lim
is solved by factoring the highest power
1
5

x 1    x 2  3  2
x

x
 lim
but
5 
x  

x 2 1  2 

x 
 lim
is solved by factoring the highest power
5
must factor “ x10 ” out of each part of the top and bottom:


5
 10 

3 
10  x

2  2 
x
5


x   x 

3 
1 
x10  2  2   9 
10 

10
x
x
x 
x 



 lim
 lim
5
x  

 x    10 
1 
1 
5
x  1  2   10 


1


10
x 
x 


x10  x  1  2 


x   1 
10


x
x 10


PL
 lim
5
5
x 2  x  x since x  
5
1

x 2   1 1  2
x

x   0  1 1  0   1
2 4 
100

x 2 1   2 
x x 

 x
 1  1
3
of “x” out of each piece:
  1 x  2  x 2
5 
 x  x  x   x  2  2 

x 
1  x  x  5  lim  
x
lim
2
x
x


x 2  2x  4
x
2
x
4
x2  2  2  2 
x
x
x 

5
1

x   1  x  1  2
x


x
 lim
2 4 
x

x 2 1   2 
x x 

2
2
2
5
1

x   1  x 2  1  2
x


x
 lim
but
2 4 
x

x 2 1   2 
x x 

77
1
5

x 1      x   3  2
x

x
5 

x 2 1  2 

x 
1
5

 x2 1   3  2
x

x   1  0  3  0   3
5 
1 0

x 2 1  2 

x 

 2  0 5  0 
1  0 5  0
74.
lim
x 


x 2  3 x  1  x is a very sneaky question. If you were
to start factoring out the highest power of “x” from each piece,
you would ultimately end up with “   0 ” which is an
indeterminate form (i.e. we have no idea what the answer might
be). We would have to start all over again!
First and foremost, we must multiply top and bottom by the
conjugate “ x 2  3x  1  x ”:
lim
x


x 2  3 x  1  x  lim
2
x  3x  1  x
x
x 2  3x  1  x

1
x 2  3x  1  x
x 2  3x  1  x
2
3x  1
 lim
x
x 2  3x  1  x
x 2  3x  1  x
Now we can factor the highest power of “x” out of each piece:
 3x 1 
x
 
3x  1
x
 x
lim
 lim
2
x

x 2  3x  1  x x  
x
3
x
1 
x2  2  2  2   x
x
x
x 

 lim
x
But:
lim
x
 lim

x 3 

3
x 1 
x
x
© 1997-2011 Grant Skene for
32
x 2  x  x since x  
1

x
1
x2
 lim
x
x
1

x 3  
x



3 1


x
1


2
x
x
x
x

 
x
x



1

x 3  
30
3
3
x




2


1 0 0 1
1 1
3 1
x  1   2  1
x x


Grant’s Tutoring (www.grantstutoring.com)
DO NOT RECOPY
SOLUTIONS TO LIMITS PRACTISE PROBLEMS (Intro Calculus)
 sin x
sin x  3 x
 lim 
x0 
2x
2x
75. lim
x0

move "2 x "
3x 

2x 
78.

 lim 
x0 
sin x
1
1
2x

Stick " x "
down here
x  1
 sin x x
3
3
   lim 

 
2 x  0  x
2 x 2



 3 h 2sin h 
3h  2sin h
sin h 

 lim 

  lim 3  2 
h
h  h  0 
h 
h0  h
h0

sin h 
lim  3  2 
 3  2 1  5
h 
h0 
1
77. lim
x 0
tan  3 x 
1

 lim tan  3 x   
x 0 
x
x
 sin  3 x 
1
 lim 
 
x  0  cos  3 x 
x
move cos  3 x 
lim
x  1
2
1

1
 x    4 x  2  since  2  1
2

2
1
1

 x    4 x  6   since 2  6  3
2

4 x2  1
lim
x  1
Balance with
"3 x" up here
 sin  3 x 
1
 lim 


x0 
1
cos  3 x 
1
x




2
lim
x  1
2
2
4 x  8x  3
2x  1

2x  3

lim
x  1
2
 2 x  1   2 x  1
 2 x  1  2 x  3 
 1
2   1
1  1 2
 2


 1
1  3 2
 1
2    3
 2
Stick " 3 x "
down here
SA
M
out of the way 
to make room for "3 x "
4 x2  8x  3
2
 1
4    2
4x  2
2  2 4
2
 lim
 2


 1
2  6
4
x  1 4 x  6
 1
2
4    6
 2
Alternate Method:
You could use ordinary factoring methods to factor the limit:
sin h
 1 [MAKE SURE YOU SAY
h
THIS!]
2
PL
But we know that lim
1
0
limit with “ x  ” as the factor on
2
0
4 x2  1
lim
x  1
1
lim
h0
is a
2
1
 1
4   1
4  1
11
0
 2
4



2
4
1

4

3
0
1
 1
 1
 1
4     8    3 4    8     3
4
2
 2
 2


 
sin h
But we know that lim
 1 [MAKE SURE YOU SAY
h0
h
THIS!]
 sin x 1 3 
1 3 4
lim 
 
 1    2
x 0 
x
2 2 
2 2 2
76.
2
4 x2  8x  3
top and bottom.
Note that when we subbed –1/2 in for x we got:
out of the way 
to make room for "x "
Balance with
" x " up here
4 x2  1
lim
E
78
 sin  3 x 
3x 
1

 lim 


x 0 
cos  3 x 
x 
3x


sin h
But we know that lim
 1 [MAKE SURE YOU SAY
h0
h
THIS!]
 sin  3 x 

1
 lim 

 3   11 3  3
x 0 
3x
cos  3 x 

1
cos 0 1
© 1997-2011 Grant Skene for
Grant’s Tutoring (text or call (204) 489-2884)
DO NOT RECOPY
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