03-limitsproblems v2

Solved Problems on Limits

and Continuity

CalculatorsMika Seppälä: Limits and Continuity

Overview of Problems

( )( )

2

0

sinlim

sinx

x

x x→

1

2

2

3 2lim

2x

x x

x→

− +−

2

3 2

3 2

1lim

3 5 2x

x x x

x x x→∞

+ + ++ + +

32 2lim 1 1

xx x

→∞+ − −

2 2lim 1 1x

x x x x→∞

+ + − − −

→ + + − − +2 20

2lim

2 1 3 1x

x

x x x x

( )0

sin 3lim

6x

x

x→

( )( )0

sin sinlimx

x

x→

4

5 6

7 8

9 10( )

( ) ( )→

+

+ + − − +2 20

2sinlim

2sin 1 sin 1x

x x

x x x x

( )tan x

2

lim ex

π→ +


Overview of Problems

11 ( )=Where tan is continuous?y x

12 ( )φφ

= − 2

1Where f sin is continuous?

1

13 ( ) ( ) −= ≠

−=

2

How must f 0 be determined so that f , 0, 1

is continuous at 0?

x xx x

x

x

14

( ) ( )

( )

− − −= = − = =

+ −

= =

2

0 0

0

Which of the following functions have removable

singularities at the indicated points?

2 8 1a) f , 2, b) g , 1

2 1

1c) h sin , 0

x x xx x x x

x x

t t tt

( ) = ∞Show that the equation sin e has many solutions.xx15


Main Methods of Limit Computations

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.

3

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.

4

In the evaluation of expressions, use the rules 2

( )0, , negativenumber .positive number

a ∞= = ∞ ∞ × = −∞

∞

The following undefined quantities cause problems: 1

0 000 , , , ,0 , .

0

∞∞∞ − ∞ ∞

∞


Main Computation Methods

If a square root appears in the expression, then multiply and divide by the conjugate of the square root expression.

3

( ) ( )

( ) ( )

1 2 1 21 2

1 2

1 2 30

1 2 1 2x

x x x x

x xx x

x x

x x x x→∞

+ − + + + −+ − − =

+ + −

+ − −= = →

+ + − + + +

Cancel out common factors of rational functions.2

( ) ( )2

1

1 111 2.

1 1 x

x xxx

x x→

− +−= = + →

− −

Use the fact that4( )

0

sinlim 1.x

x

x→=

Frequently needed rule1 ( ) ( ) 2 2.a b a b a b+ − = −


Continuity of FunctionsFunctions 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.

1

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.

4

A function f is continuous at a point x = a if2

( ) ( )limf f .x a

x a→

=

The following are not continuous x = 0:3

( ) ( ) ( )1 1f ,g sin ,h .

xx x x

x x x

= = =

Intermediate Value Theorem for Continuous Functions

Used to show that equations have solutions.


Limits by Rewriting

Problem 1

2

2

3 2lim

2x

x x

x→

− +−

Solution( )( )2 1 23 2

Rewrite 1.2 2

x xx xx

x x

− −− += = −

− −

( )2

2 2

3 2Hence lim lim 1 1.

2x x

x xx

x→ →

− += − =

−


Limits by Rewriting

Problem 23 2

3 2

1lim

3 5 2x

x x x

x x x→∞

+ + ++ + +

Solution

3 2 2 3

3 2

2 3

1 1 11

11.

3 5 23 5 21

x

x x x x x x

x x x

x x x

→∞

+ + ++ + += →

+ + + + + +


Limits by Rewriting

Problem 3 2 2lim 1 1x

x x→∞

+ − −

Solution

( )( )2 2 2 2

2 2

2 2

1 1 1 11 1

1 1

x x x xx x

x x

+ − − + + −+ − − =

+ + −

( ) ( ) ( ) ( )2 2

2 2 2 2

2 2 2 2 2 2

1 1 1 1 2

1 1 1 1 1 1

x x x x

x x x x x x

+ − − + − −= = =

+ + − + + − + + −

Rewrite

2 2

2 2

2Hence lim 1 1 lim 0.

1 1x xx x

x x→∞ →∞+ − − = =

+ + −


Limits by Rewriting

Problem 4 2 2lim 1 1x

x x x x→∞

+ + − − −

Solution

( ) ( )2 2

2 2 2 2

1 1 2 2

1 1 1 1

x x x x x

x x x x x x x x

+ + − − − += =

+ + + − − + + + − −

( )

2 2

2 22 2

2 2

1 1

1 1 1 1

1 1

x x x x

x x x xx x x x

x x x x

+ + − − − =

+ + + − −+ + − − −

+ + + − −

2 2

22

21 1 1 1

1 1x

x

x x x x

→∞

+= →

+ + + − −

Rewrite

Next divide by x.


Limits by RewritingProblem 5

→ + + − − +2 20

2lim

2 1 3 1x

x

x x x x

Solution

( )( ) ( )

( )+ + + − + + + + − += =

++ + − − +

2 2 2 2

2 2 22 2

2 2 1 3 1 2 2 1 3 1

42 1 3 1

x x x x x x x x x x

x xx x x x

( )( )( )

=+ + − − +

+ + + − +

+ + − − + + + + − +

2 2

2 2

2 2 2 2

2

2 1 3 1

2 2 1 3 1

2 1 3 1 2 1 3 1

x

x x x x

x x x x x

x x x x x x x x

( )→

+ + + − += →

+

2 2

0

2 2 1 3 11

4 x

x x x x

x

Rewrite

Next divide by x.


Limits by Rewriting

Problem 6( )

0

sin 3lim

6x

x

x→

Solution

( ) ( )sin 3 sin 31Rewrite

6 2 3

x x

x x=

( )0

sinUse the fact that lim 1.

α

α

α→=

( ) ( )0 0

sin 3 sin 3 1Since lim 1, we conclude that lim .

3 6 2x x

x x

x x→ →= =


Limits by Rewriting

Problem 7( )( )

0

sin sinlimx

x

x→

Solution

( )

( )( )( )

( )

0

0

sinsince lim 1. In the above, that fact

was applied first by substituting sin .

sin sinHence lim 1.

sinx

x

x

x

α

α

αα

→

→

=

=

=

( )( ) ( )( )( )

( )0

sin sin sin sin sin1

sin x

x x x

x x x →= →

Rewrite:


Limits by Rewriting

Problem 8( )( )

2

0

sinlim

sinx

x

x x→

Solution

( )( )

( )( )

2 2

02

sin sin1

sin sin x

x x x

x x x x →= →

Rewrite:


Limits by RewritingProblem 9 ( )

( ) ( )→

+

+ + − − +2 20

2sinlim

2sin 1 sin 1x

x x

x x x x

Solution

( )( ) ( )

( )( ) ( ) ( )( )( ) ( )( ) ( ) ( )( )

+=

+ + − − +

+ + + + − +

+ + − − + + + + − +

2 2

2 2

2 2 2 2

2sin

2sin 1 sin 1

2sin 2sin 1 sin 1

2sin 1 sin 1 2sin 1 sin 1

x x

x x x x

x x x x x x

x x x x x x x x

Rewrite

( )( ) ( ) ( )( )( )( ) ( )( )

( )( ) ( ) ( )( )( ) ( )

+ + + + − +=

+ + − − +

+ + + + − +=

− + +

2 2

2 2

2 2

2 2

2sin 2sin 1 sin 1

2sin 1 sin 1

2sin 2sin 1 sin 1

sin 2sin

x x x x x x

x x x x

x x x x x x

x x x x


Limits by Rewriting

Problem 9( )

( ) ( )→

+

+ + − − +2 20

2sinlim

2sin 1 sin 1x

x x

x x x xSolution(cont’d)

( )( ) ( )

( )( ) ( ) ( )( )( ) ( )

+

+ + − − +

+ + + + − +=

− + +

2 2

2 2

2 2

2sin

2sin 1 sin 1

2sin 2sin 1 sin 1

sin 2sin

x x

x x x x

x x x x x x

x x x x

Rewrite

( ) ( ) ( )( )( ) ( ) ( )

+ + + + − +

=

− + +

2 2sin1 2 2sin 1 sin 1

sin sinsin 2 1

xx x x x

x

x xx x

x x

→

×→ =

+0

3 22.

2 1x

Here we used the fact that all sin(x)/x terms approach 1 as x → 0.

Next divide by x.


One-sided Limits

Problem 10( )tan x

2

lim ex

π→ +

Solution

( ) ( )

( )2

tan

2

For , tan 0 and lim tan .2

Hence lim e 0.

x

x

x

x x xπ

π

ππ

→ +

→ +

< < < = −∞

=


Continuity

Problem 11 ( )Where the function tan is continuous?y x=

Solution

( ) ( )( )

( )

( )

sinThe function tan is continuous whenever cos 0.

cos

Hence tan is continuous at , .2

xy x x

x

y x x n nπ

π

= = ≠

= ≠ + ∈�


Continuity

Problem 12 ( ) 2

1Where the function f sin is continuous?

1φ

φ

= −

Solution ( ) 2

1The function f sin is continuous at all points

1

where it takes finite values.

φφ

= −

2 2

1 1If 1, is not finite, and sin is undefined.

1 1φ

φ φ

= ± − −

2 2

1 1If 1, is finite, and sin is defined and also finite.

1 1φ

φ φ

≠ ± − −

2

1Hence sin is continuous for 1.

1φ

φ

≠ ± −


Continuity

Problem 13 ( )

( )2

How must f 0 be determined so that the function

f , 0, is continuous at 0?1

x xx x x

x

−= ≠ =

−

Solution

( ) ( ) ( ) ( ) ( )

( ) ( )0

0

00

2

0 0 0

Condition for continuity of a function f at a point is:

limf f . Hence f 0 must satisfy f 0 lim f .

1Hence f 0 lim lim lim 0.

1 1

x x x

x x x

x

x x x

x xx xx

x x

→ →

→ → →

= =

−−= = = =

− −


Continuity ( )

( )

0

0

A number for which an expression f either is undefined or

infinite is called a of the function f . The singularity is

said to be , if f can be defi

singularity

removab ned in such a way le that

x x

x

0

the function f becomes continous at .x x=

Problem 14

( )

( )

( )

2

0

0

0

Which of the following functions have removable

singularities at the indicated points?

2 8a) f , 2

2

1b) g , 1

1

1c) h sin , 0

x xx x

x

xx x

x

t t tt

− −= = −

+−

= =−

= =

Answer

Removable

Removable

Not removable


Continuity

( )Show that the equation sin e has

inifinitely many solutions.

xx =

( )Observe that 0 e 1 for 0, and that sin 1 , .2

nx x n nπ

π < < < + = − ∈

�

( )

( )

Hence f 0 for if is an odd negative number 2

and f 0 for if is an even negative number.2

x x n n

x x n n

ππ

ππ

< = +

> = +

Problem 15

Solution ( ) ( ) ( ) sin e f sin e 0.x xx x x= ⇔ = − =

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.

( )We conclude that every interval 2 , 2 1 , and 0, contains 2 2

a solution of the original equation. Hence there are infinitely many solutions.

n n n nπ π

π π + + + ∈ <

�

03-limitsproblems v2

Documents