Calculus 1 Notes
Section 1.3
Page 1 of 5
Section 1.3: Computation of Limits
Big Idea: The limits of many familiar functions as x approaches a value of a is simply the function
evaluated at x = a. i.e., lim f  x   f  a 
x a
Big Skill: You should be able to compute exactly the limit of many familiar functions using the theorems from
this section.
Limit of a constant function:
For any constant c and any real number a, lim c  c
xa
Limit of a linear function:
For any real number a, lim x  a
x a
Theorem 3.1 (limits of combinations of functions)
Suppose that lim f  x  and lim g  x  both exist, and let c be any constant. Then the following apply:
xa
x a
i. lim c  f  x    c  lim f  x  
x a
 x a

(the limit of a constant times a function is the constant times the limit of the function)
ii. lim  f  x   g  x   lim f  x   lim g  x 
x a
x a
x a
(the limit of a sum of functions is the sum of the limits of each function)
iii. lim  f  x   g  x    lim f  x    lim g  x  
x a
 x a
  x a

(the limit of a product of functions is the product of the limits of each function)
f  x
 f  x   lim
 xa
iv. lim 
, provided that lim g  x   0

xa g  x 
x a
g  x

 lim
xa
(the limit of a ratio of functions is the ratio of the limits of each function)
n
n
v. lim  f  x    lim f  x   , for any positive integer n.
x a
 x a

(the limit of a power of a function is the power of the limit of the function)
Proof: You have to wait until section 1.6 …
Practice:
1. lim x 4 
x 2
2. lim3x2  2 x  4 
x 1
2 x 2  3x  5

x 0
x2  3
3. lim
Calculus 1 Notes
4. lim
x 1
Section 1.3
Page 2 of 5
x2 1

1 x
Limit of the square of a function:
Suppose that lim f  x  exists. Then,
xa
lim  f  x    lim  f  x   f  x  
x a
x a
2
 lim f  x    lim f  x 
 xa
  x a

 lim f  x  
 xa

2
Limit of a positive integer power of a function:
n
For any positive integer n, lim  f  x    lim f  x  
x a
 x a

Proof: This “follows” by mathematical induction…
n
Limit of a power function:
For any n > 0 and any real number a, lim xn  a n
xa
Proof: This “follows” from above by letting f(x) = x.
Theorem 3.2 (limit of a polynomial)
For any polynomial p(x) and any real number a, lim p  x   p  a 
x a
Proof: This “follows” from the limit of a power function and Theorem 3.1.
Theorem 3.3 (limit of roots of a function)
Suppose that lim f  x   L and n is any positive integer. Then,
x a
lim n f  x   n lim f  x 
x a
x a
. ( If n is even, then we must have that L > 0. )
 L
Proof: You have to wait until section 1.6 …
n
Calculus 1 Notes
Section 1.3
Theorem 3.4 (limits of some common transcendental functions)
For any real number a the following apply:
i. limsin  x   sin  a 
x a
ii. limcos  x   cos  a 
xa
iii. lim e x  ea
x a
iv. limln  x   ln  a  , provided that a > 0.
x a
v. limsin 1  x   sin 1  a  , provided that -1 < a < 1.
xa
vi. limcos1  x   cos1  a  , provided that -1 < a < 1.
x a
vii. lim tan 1  x   tan 1  a  , for - < a < .
xa
Proof: You have to wait until section 1.5 …
Practice:
5. lim 3 2 x 2  3x  5 
x 0
6.
lim cot x 
x  / 2
 x 1
7. lim sin 1 

x 0
 2 
8.
lim  ln  2 x   
2
x e / 2
2x

x 0 3 
x9
9. lim
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Calculus 1 Notes
Section 1.3
Page 4 of 5
Theorem 2.7 (Squeeze Theorem)
Suppose that f  x   g  x   h  x  for all x in some interval (c, d), except possibly at the point a  (c, d) and
that lim f  x   lim h  x   L for some number L. Then, lim g  x   L also.
xa
x a
x a
Picture:
The Squeeze Theorem is used to compute the limit of a function for which you can’t plug in the limiting xvalue, but for which you can find two functions that form a boundary for the function, and that have the same
limit at the given x-value.
To use the Squeeze Theorem, find two functions that form a boundary for your given function and that have the
same limit at the given x-value. Then conclude that the limit of the function is the same as the limit of the two
bounding functions.
Practice:

 1 
10. Use the Squeeze Theorem to compute lim  x 2 cos    .
x 0
 x 

Calculus 1 Notes
Section 1.3
 2 x3 
11. Use the Squeeze Theorem to prove that lim  2   0 .
x 0 x  1


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