solved problems on limits and continuity. calculators mika seppälä: limits and continuity overview...

Solved Problems on Limits and Continuity

Mika Seppälä: Limits and

ContinuityCalculators

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 1

xx 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



Overview of Problems

11 Where tan is continuous?y x

12

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

xx

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: 10 00

0 , , , , 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 21 2 3

01 2 1 2 x

x x x xx x

x xx 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

limff .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 1 111

1.3 5 23 5 2 1

x

x x x x x xx 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.

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

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:




2 20

2sinlim

2sin 1 sin 1x

x x

x x x xSolution

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




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

Solution

2

tan

2

For , tan 0 and lim tan .2

Hence lim e 0.

x

x

x

x x x



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 n



Continuity

Problem 12 2

1Where the function f sin is continuous?

1

Solution 2

1The function f sin is continuous at all points

1

where it takes finite values.

2 2

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

1 1

2 2

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

1 1

2

1Hence sin is continuous for 1.

1



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

0 0

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

21

b) g , 11

1c) h sin , 0

x xx x

xx

x xx

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 n

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

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 n

solved problems on limits and continuity. calculators mika seppälä: limits and continuity overview...

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