Top Banner
Solved Problems on Limits and Continuity
22

Solved Problems on Limits and Continuity

Mar 13, 2016

Download

Documents

Solved Problems on Limits and Continuity. Overview of Problems. 2. 1. 4. 3. 5. 6. 7. 8. 9. 10. Overview of Problems. 11. 12. 13. 14. 15. Main Methods of Limit Computations. The following undefined quantities cause problems:. 1. In the evaluation of expressions, use the rules. - PowerPoint PPT Presentation
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: Solved Problems on Limits and Continuity

Solved Problems on Limits and Continuity

Page 2: Solved Problems on Limits and Continuity

Calculators

Overview of Problems

2

0

sinlim

sinx

xx x

12

2

3 2lim2x

x xx

23 2

3 2

1lim3 5 2x

x x xx x x

3 2 2lim 1 1x

x x

2 2lim 1 1x

x x x x

2 20

2lim2 1 3 1x

xx x x x

0

sin 3lim

6x

xx

0

sin sinlimx

xx

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

Page 3: Solved Problems on Limits and Continuity

Calculators

Overview of Problems11 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 xx

x

14

2

0 0

0

Which of the following functions have removablesingularities at the indicated points?

2 8 1a) f , 2, b) g , 12 1

1c) h sin , 0

x x xx x x xx x

t t tt

Show that the equation sin e has many solutions.xx15

Page 4: Solved Problems on Limits and Continuity

Calculators

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 numbera

The following undefined quantities cause problems: 10 000 , , , , 0 , .0

Page 5: Solved Problems on Limits and Continuity

Calculators

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 xx x x x

Cancel out common factors of rational functions.2 2

11 11 1 2.1 1 x

x xx xx x

Use the fact that4 0

sinlim 1.x

xx

Frequently needed rule1 2 2.a b a b a b

Page 6: Solved Problems on Limits and Continuity

Calculators

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 ax 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.

Page 7: Solved Problems on Limits and Continuity

Calculators

Limits by RewritingProblem 1

2

2

3 2lim2x

x xx

Solution 2 1 23 2Rewrite 1.2 2

x xx x xx x

2

2 2

3 2Hence lim lim 1 1.2x x

x x xx

Page 8: Solved Problems on Limits and Continuity

Calculators

Limits by RewritingProblem 2

3 2

3 2

1lim3 5 2x

x x xx 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

Page 9: Solved Problems on Limits and Continuity

Calculators

Limits by RewritingProblem 3 2 2lim 1 1

xx 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 x

x xx x

Page 10: Solved Problems on Limits and Continuity

Calculators

Limits by RewritingProblem 4 2 2lim 1 1

xx x x x

Solution

2 2

2 2 2 2

1 1 2 2

1 1 1 1

x x x x xx x x x x x x x

2 2

2 22 2

2 2

1 1

1 1 1 11 1

x x x x

x x x xx x x xx x x x

2 2

222

1 1 1 11 1x

x

x x x x

Rewrite

Next divide by x.

Page 11: Solved Problems on Limits and Continuity

Calculators

Limits by RewritingProblem 5 2 20

2lim2 1 3 1x

xx 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

xx 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.

Page 12: Solved Problems on Limits and Continuity

Calculators

Limits by RewritingProblem 6

0

sin 3lim

6x

xx

Solution

sin 3 sin 31Rewrite 6 2 3x xx 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 xx x

Page 13: Solved Problems on Limits and Continuity

Calculators

Limits by RewritingProblem 7

0

sin sinlimx

xx

Solution

0

0

sinsince lim 1. In the above, that fact

was applied first by substituting sin .

sin sinHence lim 1.

sinx

x

xx

0

sin sin sin sin sin1

sin x

x x xx x x

Rewrite:

Page 14: Solved Problems on Limits and Continuity

Calculators

Limits by RewritingProblem 8

2

0

sinlim

sinx

xx x

Solution

2 2

02

sin sin1

sin sin x

x x xx x x x

Rewrite:

Page 15: Solved Problems on Limits and Continuity

Calculators

Limits by RewritingProblem 9

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

Page 16: Solved Problems on Limits and Continuity

Calculators

Limits by RewritingProblem 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

xx x

x xx x

0

3 2 2.2 1x

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

Next divide by x.

Page 17: Solved Problems on Limits and Continuity

Calculators

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

Page 18: Solved Problems on Limits and Continuity

Calculators

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

Page 19: Solved Problems on Limits and Continuity

Calculators

Continuity Problem 12 2

1Where the function f sin is continuous?1

Solution 2

1The function f sin is continuous at all points1

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

Page 20: Solved Problems on Limits and Continuity

Calculators

Continuity Problem 13

2

How must f 0 be determined so that the function

f , 0, is continuous at 0?1

x xx x xx

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

xx x x

x xx x xx x

Page 21: Solved Problems on Limits and Continuity

Calculators

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

singularityremovab 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 removablesingularities at the indicated points?

2 8a) f , 22

1b) g , 11

1c) h sin , 0

x xx xx

xx xx

t t tt

Answer

Removable

Removable

Not removable

Page 22: Solved Problems on Limits and Continuity

Calculators

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