2.3.3What does it look like really close?

Local Linearity

2-122.

Investigate the local behavior of $y = \frac{1}{x}$ at $x = 0.5$ by graphing the function on your calculator and zooming in.

What does the graph look like? Sketch the graph before and after you zoom in.
Since the graph of $f(x) = \frac{1}{x}$ resembles a line in a small “local” region, we say the function is locally linear. Is $f(x) = \frac{1}{x}$ truly linear close to $(0.5, 2)$ ? Why or why not?

2-123.

Study the list of basic functions below. Although the graphs vary widely by shape, some look exactly the same when you zoom in very close. What does each function look like when you zoom in at $x = –2, x = 0,$ and $x = 4$ ? Record your observations.

$y = x^2$

$y=\sin(x)$

$y = 1 - \cos(x)$

$y = x^3$

$y = \left | x \right |$

$y = \large\frac { x } { x + 1 }$

You probably noticed that $y = \left | x \right |$ does not have a local linearization at $x = 0$ because at $x = 0$ it has a cusp. What do you think the term “cusp” means and why do you think a function cannot be linearized at a cusp?

2-124.

Examine the linearization of $y=\sin(x)$ at $(0, 0)$ . Compare the values of $y=\sin(x)$ and the line $y = x$ at values close to $x = 0$ .

Complete the table below for various $x$ -values.
$\boldsymbol x$
$–1$
$–0.1$
$–0.01$
$0$
$0.01$
$0.1$
$1$
$\boldsymbol {y = \sin(x)}$
$\boldsymbol {y = x}$
For what values of $x$ is the line a good approximation of $y=\sin(x)$ ? Write a statement summarizing your findings.
On what domain is $y = x$ an overestimate? On what domain is it an underestimate? You should be able to see this information on both the graph and the table.
Estimate $\lim\limits _ { x \rightarrow 0 } \large\frac { \operatorname { sin } ( x ) } { x }$ .
Now estimate $\lim\limits _ { x \rightarrow 0 } \large\frac { 1 - \operatorname { cos } ( x ) } { x }$ .

Increasing line labeled, y = x, passing through the origin, & periodic curve labeled y = sine of x, also passing through the origin.

Math Notes

Essential Limits of Trigonometric Functions

$\lim\limits_ { h \rightarrow 0 } \frac { \operatorname { sin } ( h ) } { h } = 1$ and $\lim\limits _ { h \rightarrow 0 } \frac { 1 - \operatorname { cos } ( h ) } { h } = 0$

In the diagram at right, $\overparen{AB}$ is the arc of a circle of radius $1$ centered at $O$ , $\angle AOB$ is a small positive angle of measure $h$ (radians) with $\overline{BC}$ and $\overline{DA}$ perpendicular to $\overline{OA}$ .

Right triangle, D, O, A, with vertical leg, A D, & horizontal leg, O A, angle opposite vertical leg labeled, h, point B on hypotenuse, close to D, point C on horizontal leg, such that segment, C B, is vertical, with arc between point, a, & point, b.

Proof 1:

Since $BC=\sin(h),OC=\cos(h)$ , and $AD=\tan(h),$ and since area $(\Delta OCB)<\text{ area (sector }OAB)<\text{ area }(\Delta OAD),$ we can write $\frac { 1 } { 2 }\sin(h) \cos(h) < \frac { 1 } { 2 } h < \frac{1}{2}\tan(h)$ . Dividing by $\frac { 1 } { 2 }\sin(h),$ we have $\cos(h)<\frac{\sin\left(h\right)}{h}<\frac{1}{\cos(h)}$ . Taking reciprocals, $\frac { 1 } { \operatorname { cos } ( h ) } > \frac { h } { \operatorname { sin } ( h ) } > \cos(h)$ . Since $\lim\limits _ { h \rightarrow 0 ^ { + } } \operatorname { cos } ( h ) = \operatorname { lim } _ { h \rightarrow 0 ^ { + } } \frac { 1 } { \operatorname { cos } ( h ) } = 1$ and since $\frac { \operatorname { sin } ( h ) } { h }$ is between the two values for all $h > 0$ , the limit of $\frac { \operatorname { sin } ( h ) } { h }$ is “squeezed” between the limits and must equal $1$ . Since $\frac { \operatorname { sin } ( h ) } { h }$ is an even function, we also see that $\lim\limits _ { h \rightarrow 0 ^ { - } } \frac { \operatorname { sin } ( h ) } { h } = 1$ . Thus, $\lim\limits _ { h \rightarrow 0 } \frac { \operatorname { sin } ( h ) } { h } = 1$ .

Proof 2:

$\frac { 1 - \operatorname { cos } ( h ) } { h } = \frac { 1 - \operatorname { cos } ^ { 2 } ( h ) } { h ( 1 + \operatorname { cos } ( h ) ) } = \frac { \operatorname { sin } ^ { 2 } ( h ) } { h ( 1 + \operatorname { cos } ( h ) ) } = \frac { \operatorname { sin } ( h ) } { h } \cdot \frac { \operatorname { sin } ( h ) } { 1 + \operatorname { cos } ( h ) }$ .

Hence, $\lim\limits _ { h \rightarrow 0 } \frac { 1 - \operatorname { cos } ( h ) } { h } = \lim\limits _ { h \rightarrow 0 } \frac { \operatorname { sin } ( h ) } { h } \cdot \lim\limits _ { h \rightarrow 0 } \frac { \operatorname { sin } ( h ) } { 1 + \operatorname { cos } ( h ) }$ .

By (1), the first limit is $1$ , and the second limit is $\frac { 0 } { 2 } = 0$ . Hence $\lim\limits _ { h \rightarrow 0 } \frac { 1 - \operatorname { cos } ( h ) } { h } = 0$ .

Review and Preview problems below

2-125.

Determine if the following functions are even, odd, or neither. Explain how you determined your choice. Homework Help ✎

$y = \sin2(x)$

$y = \large\frac { x ^ { 2 } + 1 } { x ^ { 3 } - 2 x }$

2-126.

The rate of customers who pass through the checkout stand at a grocery store depends on the time of day. Assume the rate follows the following piecewise-defined function, where $x$ represents the time of day (in hours) after 9:00 a.m. and $f (x)$ is the number of customers served per hour. Homework Help ✎

$f ( x ) = \left\{ \begin{array} { l l } { 120 x } & { \text { for } 0 \leq x < 2 } \\ { 180 x - 120 } & { \text { for } 2 \leq x < 5 } \\ { 240 x - 420 } & { \text { for } x \geq 5 } \end{array} \right.$

Graph the function and state its domain and range.
Is this function continuous?
Calculate the area under the curve for $0\le x\le8$ . What are the units of this area? What does this area represent?

2-127.

Fertilizer needs to be applied during the fastest growth of the plant. At right is the graph of the growth cycle of a flowering shrub. Homework Help ✎

Using complete sentences, write a detailed statement describing the growth of this shrub for $0\le t\le5$ months.
During what time interval should the plant be fertilized?
Approximately how fast is the shrub growing at $t = 3$ ? How did you get your answer?
What is the shrub's average rate of growth over the complete growth cycle? How did you get your answer?

2-128.

Rotate the flag shown at right (created by the region under the curve for $0\le x\le4$ ) about the $x$ -axis. Describe (and draw) the shape that is created and calculate its volume. 2-128 HW eTool. Homework Help ✎

First quadrant with enclosed shaded polygon, starting at the origin, up 2, right 1, up 1, right 1, down 2, right 1, up 1, right 1, down 2, left 4, back to the origin to enclose the figure.

2-129.

What is the relationship between the slopes of perpendicular lines? If you know the slope of one of the lines, how can you determine the slope of the perpendicular line? Homework Help ✎

2-130.

A function, $f$ , is continuous for all real numbers. If $f(x) = \large\frac { x ^ { 2 } - 9 } { x + 3 }$ when $x\ne-3$ , then what must $f(-3)$ equal? Write a piecewise-defined function that represents this situation. Homework Help ✎

2-131.

Evaluate the following limits. Homework Help ✎

$\lim\limits _ { x \rightarrow \infty }\large \frac { \sqrt { 9 x ^ { 2 } - 3 x + 1 } } { 4 x ^ { 2 } - 1 }$

$\lim\limits _ { x \rightarrow - 1 } \large\frac { x + 1 } { x ^ { 2 } + 5 x + 6 }$

$\lim\limits _ { x \rightarrow - \infty }\large \frac { 2 - 3 x - 4 x ^ { 2 } } { ( 1 - 3 x ) ^ { 2 } }$

$\lim\limits _ { x \rightarrow 2 } \large\frac { | x ^ { 2 } - 4 | } { x - 2 }$

$\boldsymbol x$	$–1$	$–0.1$	$–0.01$	$0$	$0.01$	$0.1$	$1$
$\boldsymbol {y = \sin(x)}$
$\boldsymbol {y = x}$

Calculus Third Edition

Essential Limits of Trigonometric Functions

Local Linearity