Homework Homework Assignment #3 Read Section 2.4 Page 91, Exercises: 1 – 33 (EOO) Rogawski Calculus Copyright © 2008 W. H. Freeman and Company

Homework

Homework Assignment #3 Read Section 2.4 Page 91, Exercises: 1 – 33 (EOO)

Rogawski CalculusCopyright © 2008 W. H. Freeman and Company

Homework, Page 82Evaluate the limits using the Limit Laws and the following where c and k are constants.


91. lim

xx

9lim 9x

x



3

5. lim 3 4x

x

3

3

lim 3 4 3 3 4 9 4 5

lim 3 4 5

x

x

x

x



4

9. lim 3 14t

t

4

4

lim 3 14 3 4 14 12 14 2

lim 3 14 2

t

t

t

t



2

13. lim 1 2x

x x x

2

2

lim 1 2 2 2 1 2 2 2 3 4 24

lim 1 2 24

x

x

x x x

x x x



3

117. lim

1x

x

x

3

3

1 1 3 2 1lim

1 1 3 4 2

1 1lim

1 2

x

x

x

x

x

x



2 3

321. lim 9

xx x

2 12 3

3

2 3

3

1 27 1 28lim 9 3 9 3 9 9

27 3 3 3

28lim 9

3

x

x

x x

x x

Homework, Page 82Evaluate the limit assuming:


4

25. limx

f x g x

4 4

lim 3, lim 1x x

f x g x

4

4

lim 3 1 3

lim 3

x

x

f x g x

f x g x

Homework, Page 8229. Can the Quotient Law be used to evaluate : ?

Explain.

If the limit is rewritten as:

using the Product Law, we obtain which does not

exist, resulting in a nonexistent limit. Using the Quotient Law, direct substitution yields an indeterminate form, resulting in another nonexistent limit.


0

sinlimx

x

x

0 0 0

sin 1lim limsin limx x x

xx

x x

0

1limx x

Homework, Page 8233. Use the Limit Laws and the result to show that for all whole numbers.


limx c

x c

lim n n

x cx c

1 11

1 11

lim lim lim lim

lim

: lim lim

: lim

: lim lim

n n

x c x c x c x c

n n

x c

x c x c

k kk x c

k k k kk x c x c

x c x x x c c c

x c

P x c x c

P x c

P x x c c x c


Jon Rogawski

Calculus, ET First Edition

Chapter 2: LimitsSection 2.4: Limits and Continuity


If the graph of a function may be drawn without lifting the pencil from the page, the graph and thus the function aresaid to be continuous , such as at x = c in Figure 1. A break in the graph, such as at x = c in Figure 2, is called a discontinuity.


The discussion on the previous slide leads to the following definition:

lim , lim , and

lim for all on [ , ].x a x b

x c

f x f a f x f b

f x f c c a b

If a continuous function is defined on [a, b], and c is any point on (a, b) then the following conditions exist:


Three conditions must hold for a function to be continuous: 1. lim exists

x cf x

2. existsf c

3. limx c

f x f c

Figures 3 and 4 are graphs of continuous functions.


Some functions contain discontinuities that can be removed by redefining the piecewise function, such as defining f (2) = 5 instead of f (2) = 10 for the function graphed in Figure 5. Such discontinuities are called removable discontinuities.


A jump discontinuity is a discontinuity for which the first condition of continuity is not met, that is:

lim limx c x c

f x f x

lim or limx c x c

f x f c f x f c

If either , the function is said to be one-sided continuous as defined below:


A jump discontinuity may be one-sided continuous as in Figure 6 (A) or neither right– nor left–continuous as in 6(B).

A function with a jump discontinuity may not be both right–and left–continuous at the jump discontinuity, as it would no longer pass the vertical line test, meaning it is no longera function.


An infinite discontinuity is one in which one or both one-sided limits are infinite. Since infinity is a concept rather than an actual number, f (2) is not defined in Figure 8 (A) or 8 (B). Some texts state that the limit at x = 2 in Figure8 (A) does not exist, although ours does not so state.


Figure 9 illustrates what some refer to as an oscillating discontinuity. In this case, neither the right– nor left–sided limits exist at x = 0, so neither does the limit.

Example, Page 912. Find the points of discontinuity and state whether f (x) is left- or right-continuous, or neither at these points. At which point does f (x) have a removable discontinuity?


x

y


Functions “built” of functions known to be continuous are also continuous in accordance with Theorem 1.

Theorem 1 is proven as follows:


As shown in the proof of Theorem 2, all polynomials are continuous.


Figure 10 shows the graphs of some basic functions that arecontinuous on their domains. This is true of most basic functions.


Theorem 3 formally states the continuity of some basic functions.

Since inverse functions are reflections of the parent function about the function y = x, the inverses of continuous functions are also continuous on their domains, as stated in Theorem 4.

Example, Page 91Use the Laws of Continuity and Theorems 2–3 to show that the function is continuous.


2 cos

12. 3 cos

x xf x

x

Example, Page 91Determine the points at which the function is discontinuous and state the type of discontinuity.


220.

2

xf x

x



3 3226. 3 9g t t t



34. f x x x


Just as products and quotients of continuous functions are continuous on their domains, compositions of continuous functions are also continuous as stated in Theorem 5.


When a function f (x) is known to be continuous, then we may state . This method of evaluating a limit is sometimes called the substitution method as the value of c is substituted for x in the function to obtain the limit. The substitution method can not be used on the greatest integer function shown in Figure 12 as it is not everywhere continuous.

limx c

f x f c


Sometimes continuous functions are used to model physicalquantities that on a large-scale may mimic a continuous function, but on a small scale these quantities are incremental and thus discontinuous. Figure 13 shows twosuch functions.

Homework

Homework Assignment #4 Read Section 2.5 Page 91, Exercises: 1 – 33 (EOO) Quiz next time


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Homework Homework Assignment #3 Read Section 2.4 Page 91, Exercises: 1 – 33 (EOO) Rogawski Calculus Copyright © 2008 W. H. Freeman and Company