Solved Problems on Limits
and Continuity
Overview of Problems
1
x  3x  2
lim
x 2
x 2
2
x3  x2  x  1
lim
x  x 3  3 x 2  5 x  2
3
lim x 2  1  x 2  1
4
lim x 2  x  1  x 2  x  1
2
5
x 
2x
lim
2x  x  1  x  3 x  1
x 0
7
lim
2
sin  sin  x  
x 0
9
2
lim
x 0
Calculators
8
x 
6
lim
x 0
lim
sin  3 x 
x 0
6x
 
sin x 2
x sin  x 
x
x  2sin  x 
x  2sin  x   1  sin  x   x  1
2
2
10
Mika Seppälä: Limits and Continuity
lim e

x 
2
tan x 
Overview of Problems
11
12
13
14
Where y  tan  x  is continuous?
 1 
Where f    sin  2
 is continuous?


1


x2  x
How must f  0  be determined so that f  x  
, x  0,
x 1
is continuous at x  0?
Which of the following functions have removable
singularities at the indicated points?
15
a)
x 2  2x  8
f x 
, x0  2, b)
x2
c)
 1
h  t   t sin   , t0  0
t 
g x  
x 1
, x0  1
x 1
Show that the equation sin  x   ex has  many solutions.
Calculators
Mika Seppälä: Limits and Continuity
Main Methods of Limit Computations
1
2
The following undefined quantities cause problems:
0 
00 , , ,   ,0 , 0.
0 
In the evaluation of expressions, use the rules
a

 0,
 ,   negative number   .

positive number
3
4
If the function, for which the limit needs to be computed, is
defined by an algebraic expression, which takes a finite value
at the limit point, then this finite value is the limit value.
If the function, for which the limit needs to be computed,
cannot be evaluated at the limit point (i.e. the value is an
undefined expression like in (1)), then find a rewriting of the
function to a form which can be evaluated at the limit point.
Calculators
Mika Seppälä: Limits and Continuity
Main Computation Methods
1
2
3
Cancel out common factors of rational functions.
x 2  1  x  1  x  1

 x  1 
 2.
x 1
x 1
x 1
If a square root appears in the expression, then multiply and
divide by the conjugate of the square root expression.
x 1 
4
 a  b a  b  a2  b2.
Frequently needed rule

x 2 
x 1  x 2
x 1 
 x  1   x  2 

x 1  x 2
Use the fact that lim
x 0
Calculators

sin  x 
x
x 1 
x 2

x 2
3
x 1 
 1.
Mika Seppälä: Limits and Continuity
x 2

0
x 
Continuity of Functions
1
2
Functions defined by algebraic or elementary expressions
involving polynomials, rational functions, trigonometric
functions, exponential functions or their inverses are
continuous at points where they take a finite well defined
value.
A function f is continuous at a point x = a if
limf  x   f  a .
x a
3
4
The following are not continuous x = 0:
1
x
1
f  x   , g  x   sin   ,h  x  
.
x
x
x
Used to show
that equations
have solutions.
Intermediate Value Theorem for Continuous Functions
If f is continuous, f(a) < 0 and f(b) > 0, then there is a point
c between a and b so that f(c) = 0.
Calculators
Mika Seppälä: Limits and Continuity
Limits by Rewriting
Problem 1
Solution
x 2  3x  2
lim
x 2
x 2
x 2  3 x  2  x  1 x  2 
Rewrite

 x  1.
x 2
x 2
x 2  3x  2
Hence lim
 lim  x  1  1.
x 2
x 2
x 2
Calculators
Mika Seppälä: Limits and Continuity
Limits by Rewriting
Problem 2
x3  x2  x  1
lim 3
x  x  3 x 2  5 x  2
Solution
1

3
2
x  x  x 1
x

x 3  3x 2  5x  2 1 3 
x
1
Calculators
1
1

x 2 x 3 1.
x 
5
2
 3
2
x
x
Mika Seppälä: Limits and Continuity
Limits by Rewriting
lim x 2  1  x 2  1
Problem 3
x 
Solution
Rewrite
x 1 x
2


2

1 
 
2
x 1 
2

x2  1  x2  1
x2  1  x2  1

x 1
2
x2  1  x2  1
2
x


2
 
x 
Calculators

 1  x2  1
x2  1  x2  1
Hence lim x  1  x  1  lim
2

x2  1  x2  1
x2  1  x2  1
2
2
x 
2
x 1 x 1
2
Mika Seppälä: Limits and Continuity
2
 0.
Limits by Rewriting
lim x 2  x  1  x 2  x  1
Problem 4
x 
Solution
Rewrite
x2  x  1  x2  x  1 

x


2

x  x 1 x  x 1
2
2
 

x2  x  1  x2  x  1
x2  x  1  x2  x  1
 x  1  x2  x  1
x2  x  1  x2  x  1
2

1
Calculators
2
x
1 1
1 1
 2  1  2
x x
x x
2x  2
x2  x  1  x2  x  1

2
x 
Mika Seppälä: Limits and Continuity
Next divide by x.
Limits by Rewriting
Problem 5
2x
lim
2x 2  x  1  x 2  3 x  1
x 0
Solution
Rewrite
2x

2x  x  1  x  3 x  1
2
2
2x



2x 2  x  1  x 2  3 x  1

2x 2  x  1  x 2  3 x  1
2x 



 2x  x  1   x  3x  1
2  2x  x  1  x  3 x  1


1
2x

2x 2  x  1  x 2  3 x  1
2
2
2
2
2
x4
Calculators
2

2x 2  x  1  x 2  3 x  1
x 0
Mika Seppälä: Limits and Continuity

2x 2  x  1  x 2  3 x  1
x 2  4x
Next divide by x.
Limits by Rewriting
Problem 6
lim
x 0
Solution
Rewrite
sin  3 x 
6x
Use the fact that lim
 0
sin  3 x 
Since lim
6x
Calculators
3x

 1.
1 sin  3 x 

2 3x
sin  3 x 
x 0
sin  
 1, we conclude that lim
x 0
Mika Seppälä: Limits and Continuity
sin  3 x 
6x
1
 .
2
Limits by Rewriting
Problem 7
lim
sin  sin  x  
x 0
Solution
x
Rewrite:
sin  sin  x  
x
since lim

sin  sin  x   sin  x 
sin  x 
sin  
 0

x

1
x 0
 1. In the above, that fact
was applied first by substituting   sin  x .
Hence lim
x 0
Calculators
sin  sin  x  
sin  x 
 1.
Mika Seppälä: Limits and Continuity
Limits by Rewriting
 
sin x 2
Problem 8
lim
Solution
Rewrite:
x 0
x sin  x 
   sin  x 
sin x 2
x sin  x 
Calculators
2
x2
x

1
x 0
sin  x 
Mika Seppälä: Limits and Continuity
Limits by Rewriting
Problem 9
x  2sin  x 
lim
x 2  2sin  x   1  sin2  x   x  1
x 0
Solution
Rewrite
x  2sin  x 
x  2sin  x   1  sin  x   x  1
2

2
 x  2sin  x   


x 2  2sin  x   1  sin2  x   x  1


Calculators
 x  2sin  x   
x
2

x 2  2sin  x   1  sin2  x   x  1

x 2  2sin  x   1  sin2  x   x  1
 

 2sin  x   1  sin2  x   x  1
 x  2sin  x   

x 2  2sin  x   1  sin2  x   x  1

x 2  2sin  x   1  sin2  x   x  1
x 2  sin2  x   2sin  x   x
Mika Seppälä: Limits and Continuity
Limits by Rewriting
Solution
(cont’d)
Problem 9
lim
x 0
x  2sin  x 
x 2  2sin  x   1  sin2  x   x  1
Rewrite
x  2 sin  x 
x 2  2 sin  x   1  sin2  x   x  1

 x  2 sin  x   
x 2  2 sin  x   1  sin2  x   x  1
x  sin  x   2 sin  x   x
2

Next divide by x.
2

sin  x  
2
2
1

2

 x  2sin  x   1  sin  x   x  1
x 

sin  x 
sin  x 
x  sin  x 
2
1
x
x




x 0
32
 2.
2 1
Here we used the fact that all sin(x)/x terms approach 1 as x  0.
Calculators
Mika Seppälä: Limits and Continuity
One-sided Limits
lim e
Problem 10
tan x 

x 
2
Solution
For

2
 x   , tan  x   0 and lim tan  x   .

x 
2
Hence lim e

tan x 
 0.
x 
2
Calculators
Mika Seppälä: Limits and Continuity
Continuity
Problem 11
Where the function y  tan  x  is continuous?
Solution
The function y  tan  x  
sin  x 
cos  x 
is continuous whenever cos  x   0.
Hence y  tan  x  is continuous at x 
Calculators

2
 n , n  .
Mika Seppälä: Limits and Continuity
Continuity
Problem 12
Solution
 1 
Where the function f    sin  2
 is continuous?
   1
 1 
The function f    sin  2
 is continuous at all points
   1
where it takes finite values.
 1 
1
is
not
finite,
and
sin
 2
 is undefined.
2
 1
   1
 1 
1
If   1, 2
is finite, and sin  2
 is defined and also finite.
 1
   1
If   1,
 1 
Hence sin  2
 is continuous for   1.
   1
Calculators
Mika Seppälä: Limits and Continuity
Continuity
Problem 13
How must f  0  be determined so that the function
x2  x
f x 
, x  0, is continuous at x  0?
x 1
Solution
Condition for continuity of a function f at a point x0 is:
lim f  x   f  x0  . Hence f  0  must satisfy f  0   lim f  x  .
x  x0
x 0
x  x  1
x2  x
Hence f  0   lim
 lim
 lim x  0.
x 0 x  1
x 0
x 0
x 1
Calculators
Mika Seppälä: Limits and Continuity
Continuity
A number x0 for which an expression f  x  either is undefined or
infinite is called a singularity of the function f . The singularity is
said to be removable, if f  x0  can be defined in such a way that
the function f becomes continous at x  x0.
Problem 14
Which of the following functions have removable
singularities at the indicated points?
a)
b)
c)
Calculators
Answer
x 2  2x  8
f x 
, x0  2
x2
x 1
g x  
, x0  1
x 1
Removable
 1
h  t   t sin   , t0  0
t 
Removable
Mika Seppälä: Limits and Continuity
Not removable
Continuity
Problem 15
Show that the equation sin  x   e x has
inifinitely many solutions.
Solution
sin  x   ex  f  x   sin  x   ex  0.
By the intermediate Value Theorem, a continuous function takes
any value between any two of its values. I.e. it suffices to show
that the function f changes its sign infinitely often.
n


Observe that 0  e x  1 for x  0, and that sin   n    1 , n  .
2


Hence f  x   0 for x   n if n is an odd negative number
2

and f  x   0 for x   n if n is an even negative number.
2



We conclude that every interval   2n ,   2n  1  , n  and n  0, contains
2
2

a solution of the original equation. Hence there are infinitely many solutions.
Calculators
Mika Seppälä: Limits and Continuity