Homework3—Due9/20/2010inclass Name: Solutions ...

Homework 3 -- Due 9/20/2010 in class

Name: Solutions

Section Number: all

This cover sheet must be attached as the top page of your homework. 1. Given the function f (x) defined by the following graph:

(a) Evaluate the following limits if they exist: limx4 f (x), limx1- f (x), limx1+ f (x), limx1 f (x).

(b) Is f (x) continuous at the point x = -4? Explain why or why not. (c) Is f (x) continuous at the point x = 1? Explain why or why not. (d) Is f (x) continuous on the interval (-4, 1)? Explain why or why not. (e) Is f (x) continuous on the interval [-4, 1]? Explain why or why not.

2. Evaluate the limit:

3. Evaluate the limit:

sin 4x

lim

.

x0 9x

x2 cos 2x

lim

.

x0 1 - cos x

4. Find a constant k so that f (x) is continuous for all x:

x2 - 1, x < 3

f (x) =

.

2kx, x 3

5. Show (using one of the theorems from this class) that the following equation has at least one

solution on the given interval. Clearly state the theorem you are using and explain why it

applies in this case:

cos x = x2 - 1 on (0, ).

6. Show using the formal definition of the limit (epsilon-delta definition) that

lim (4x - 2) = 10.

x3

Copyright 2010, Victoria Howle and Department of Mathematics and Statistics, Texas Tech University. No part of this document may be reproduced, redistributed, or transmitted in any manner without the permission of the instructor.

1. Given the function f (x) defined by the following graph:

(a) Evaluate the following limits if they exist:

limx4 f (x),

limx1- f (x),

limx1+ f (x),

limx1 f (x).

lim f (x) 1 (left and right limits both go to 1)

x4

lim f (x) = 2

x1-

lim f (x) = 4

x1+

lim f (x)

x1

does not exist since left and right limits are not equal

(Note that in the first limit, I meant to ask about the limit as x goes to -4; that limit also exists since the left and right limits both exist and equal -2.)

(b) Is f (x) continuous at the point x = -4? Explain why or why not. Yes, f (x) is continuous at x = -4. We have three things to show: (1) f (-4) is defined and equals approximately -2. (2) The limit exists as x -4.

lim f (x) = -2 (left limit)

x-4-

lim f (x) = -2 (right limit)

x-4+

therefore lim f (x) = -2

x-4

(3) f (-4) = -2 = limx-4 f (x) Therefore the function is continuous at this point.

(c) Is f (x) continuous at the point x = 1? Explain why or why not. No. The limit from the left goes to 2 and the limit from the right goes to 4, so the two-sided limit does not exist at x = 1.

(d) Is f (x) continuous on the interval (-4, 1)? Explain why or why not. Yes. There are no suspicious points in this interval. The function is continuous as every point in the open interval (-4, 1).

(e) Is f (x) continuous on the interval [-4, 1]? Explain why or why not. No. We already know that f (x) is continuous on the open interval (-4, 1), so we need to check continuity from the left and right at the end points.

lim f (x) = -2 = f (-4) so f(x) is continous from the right at -4

x-4+

lim f (x) = 2 = f (1) so f(x) is not continous from the left at 1

x1-

f (x) is not continuous from the left at x = 1, so it is not continuous on the closed interval [-4, 1]. Note that it is, however, continuous on the half-open interval [-4, 1).

Copyright 2010, Victoria Howle and Department of Mathematics and Statistics, Texas Tech University. No part of this document may be reproduced, redistributed, or transmitted in any manner without the permission of the instructor.

2. Evaluate the limit:

sin 4x

lim

.

x0 9x

We

want

to

make

this

look

like

the

limit

we

know:

limx0

sin u u

=

1

sin 4x

sin 4x 4

lim

= lim

?

x0 9x

x0 4x 9

4 =

9

3. Evaluate the limit:

x2 cos 2x

lim

.

x0 1 - cos x

x2 cos 2x

x2 cos 2x 1 + cos x

lim

= lim

?

x0 1 - cos x

x0 1 - cos x 1 + cos x

x2 cos 2x + x2 cos 2x cos x

= lim

x0

1 - cos2 x

x2 cos 2x + x2 cos 2x cos x

= lim

x0

sin2 x

x2 cos 2x

x2 cos 2x cos x

=

lim

x0

sin2 x

+ lim

x0

sin2 x

x

x

x

x

= lim

? lim

? lim cos 2x + lim

? lim

? lim cos 2x cos x

x0 sin x x0 sin x x0

x0 sin x x0 sin x x0

= 1+1

=2

Copyright 2010, Victoria Howle and Department of Mathematics and Statistics, Texas Tech University. No part of this document may be reproduced, redistributed, or transmitted in any manner without the permission of the instructor.

4. Find a constant k so that f (x) is continuous for all x:

x2 - 1, x < 3

f (x) =

.

2kx, x 3

Both parts of this piecewise-defined function are polynomials, so we only need to check continuity at the suspicious point where the function definition changes (i.e., at x = 3). For continuity at x = 3, we need for the limit to exist (left and right limits must exist and be equal) at x = 3 and for that limit to equal the value f (3). We'll do the left limit first since there are no k s in that limit and its value will determine how we need to set k:

lim f (x) = lim x2 - 1 = 8.

x3-

x3-

We need to set k so that the limit from the right and the function value are both equal to 8

at x = 3:

lim f (x) = lim 2kx = 6k.

x3+

x3+

We need this to equal the left limit, so we must set k = 8/6 = 4/3. Then we have

8

lim f (x) = lim x = 8.

x3+

x3+ 3

This also gives us f (3) = 8, therefore with k = 4/3, the limit at the suspicious point exists and equals the function evaluated at the point. Thus with this choice of k the function is continuous at x = 3 and therefore for all x.

Copyright 2010, Victoria Howle and Department of Mathematics and Statistics, Texas Tech University. No part of this document may be reproduced, redistributed, or transmitted in any manner without the permission of the instructor.

5. Show (using one of the theorems from this class) that the following equation has at least one solution on the given interval. Clearly state the theorem you are using and explain why it applies in this case:

cos x = x2 - 1 on (0, ).

We'll use the Root Location Theorem: If f is continuous on the closed interval [a, b] and if f (a) and f (b) have opposite signs, then there exists at least one point x = c on the open interval (a, b) such that f (c) = 0. In this case, the given equation has a solution if cos x - x2 + 1 = 0, so our function is f (x) = cos x - x2 + 1. The function is a difference of two functions that are continuous over all real numbers (cosine and a polynomial), so is continuous over all real numbers, thus it's continuous over the closed interval [0, ]. Checking values at the endpoints, we have f (0) = cos 0 + 1 = 2 > 0 and f () = cos - 2 + 1 = -1 - 2 + 1 = -2 < 0. Therefore, there is at least one point c such that f (c) = 0, and therefore the equation has at least one solution on the given interval. (Note that we could also have chosen to use the Intermediate Value Theorem to prove this result.)

6. Show using the formal definition of the limit (epsilon-delta definition) that

lim (4x - 2) = 10.

x3

Proof. We need to show that given an > 0 we can find a > 0 such that |x - 3| < implies |f (x) - 10| < . Let > 0 be given. We want to show that |f (x) - 10| < , i.e.,

|f (x) - 10| < |4x - 2 - 10| <

|4x - 12| < 4|x - 3| < |x - 3| < 4

So we want to choose = 4 . I.e., given an > 0, let = 4 . Then,

|x - 3| < |x - 3| <

4 4|x - 3| < |4x - 12| < |(4x - 2) - 10| < |f (x) - 10| <

Copyright 2010, Victoria Howle and Department of Mathematics and Statistics, Texas Tech University. No part of this document may be reproduced, redistributed, or transmitted in any manner without the permission of the instructor.

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