Chapter 13.1, Problem 34E

a.

To determine

To evaluate : The function $h\left(x\right)$.

Answer to Problem 34E

$x$	$1$	$0.5$	$0.1$	$0.05$	$0.01$	$0.005$
$h\left(x\right)$	$0.55740772$	$0.37041992$	$0.33467209$	$0.333667$	$0.33334667$	$0.33333667$

Explanation of Solution

Given information:

  $h\left(x\right)=\frac{\tan x-x}{x^3}$

  $x=1$ , $0.5$ , $0.1$ , $0.05$ , $0.01$ and $0.005$

Concept used:

We substitute the value of $x$ to the right hand of the equation and evaluate.

Calculation:

  $h\left(x\right)=\frac{\tan x-x}{x^3}$

$x$	$1$	$0.5$	$0.1$	$0.05$	$0.01$	$0.005$
$h\left(x\right)$	$0.55740772$	$0.37041992$	$0.33467209$	$0.333667$	$0.33334667$	$0.33333667$

b.

To determine

To guess : The value of the given function.

Answer to Problem 34E

  $\underset{x\to 0}{\lim }h\left(x\right)=0.3333$

Explanation of Solution

Given information:

  $\underset{x\to 0}{\lim }\frac{\tan x-x}{x^3}$

Formula used:

If $f\left(x\right)$ is a function and the value of $f\left(x\right)$ approaches $L$ as $x$ approaches $a$ from the right then, $\underset{x\to a^{+}}{\lim }f\left(x\right)=L$

If $f\left(x\right)$ is a function and the value of $f\left(x\right)$ approaches $L$ as $x$ approaches $a$ from the left then, $\underset{x\to a^{-}}{\lim }f\left(x\right)=L$

  $\underset{x\to a}{\lim }f\left(x\right)=L\text{ if and only if }\underset{x\to a^{-}}{\lim }f\left(x\right)=L\text{ and }\underset{x\to a+}{\lim }f\left(x\right)=L$

If the left-hand and right-hand limits are different, the limit does not exist.

Calculation:

Let, $h\left(x\right)=\frac{\tan x-x}{x^3}$

$x$	$1$	$0.5$	$0.1$	$0.05$	$0.01$	$0.005$
$h\left(x\right)$	$0.55740772$	$0.37041992$	$0.33467209$	$0.333667$	$0.33334667$	$0.33333667$

$x$	$-1$	$-0.5$	$-0.1$	$-0.05$	$-0.01$	$-0.005$
$h\left(x\right)$	$0.55740772$	$0.37041992$	$0.33467209$	$0.333667$	$0.33334667$	$0.33333667$

From the above table we can conclude,

  $\underset{x\to 0^{-}}{\lim }h\left(x\right)=0.3333$ ; $\underset{x\to 0^{+}}{\lim }h\left(x\right)=0.3333$

Therefore, $\underset{x\to 0}{\lim }h\left(x\right)=0.3333$

c.

To determine

To evaluate : The $h\left(x\right)$ until it gives the value $0$ and to tell whether the guess in the part (b) is correct or not and why I get the value of $h\left(x\right)$ as $0$

Answer to Problem 34E

$x$	$0.005$	$0.0001$	$0.00001$	$0.000001$	$\text{0}\text{.0000001}$	$0.00000001$
$h\left(x\right)$	$0.33333667$	$0.33333334$	$0.33333288$	$0.33330746$	$0.33087225$	$0$

Yes, the guess in the (b) is correct.

The value of $h\left(x\right)$ is obtained as $0$ because the value of numerator becomes smaller as the value of $x$ smaller. The calculator has the certain limit of number to be calculated. That is why, the calculator shows the value of the function $0$ for very small value of $x$.

Explanation of Solution

Given information:

  $h\left(x\right)=\frac{\tan x-x}{x^3}$

Formula used:

We substitute the value of $x$ to the right hand of the equation and evaluate.

Calculation:

  $h\left(x\right)=\frac{\tan x-x}{x^3}$

$x$	$0.005$	$0.0001$	$0.00001$	$0.000001$	$\text{0}\text{.0000001}$	$0.00000001$
$h\left(x\right)$	$0.33333667$	$0.33333334$	$0.33333288$	$0.33330746$	$0.33087225$	$0$

Yes, the guess in the (b) is correct.

The value of $h\left(x\right)$ is obtained as $0$ because the value of numerator becomes smaller as the value of $x$ smaller. The calculator has the certain limit of number to be calculated. That is why, the calculator shows the value of the function $0$ for very small value of $x$.

d.

To determine

To graph : The function $h\left(x\right)$ in the viewing rectangle $\left[-1,1\right]\text{ and }\left[0,1\right]$ . To get the distortion at the specified portion by zooming the graph and to compare the result with part(c).

Explanation of Solution

Given information:

  $h\left(x\right)=\frac{\tan x-x}{x^3}$

Concept used:

Graph is sketched using graphing calculator.

Graph:

  $h\left(x\right)=\frac{\tan x-x}{x^3}$

  

The above graph of $h\left(x\right)=\frac{\tan x-x}{x^3}$ represented in the viewing rectangle $\left[-1,1\right]\text{ and }\left[0,1\right]$.

  

The above graph of $h\left(x\right)=\frac{\tan x-x}{x^3}$ shows the distortion.

In part (c) we get the value of $h\left(x\right)$ near to $0.3333$ as $x$ approaches $0$ . But when the value of $x$ become very small then it shows $0$ . Due to limitation of the calculator it shows $0$ and for the same reason, this distortion is showing.