Chapter 2.5, Problem 12E

a.

To determine

To find: the given limit by evaluating the given function for value of x that approaches from the left and from the right.

Answer to Problem 12E

  $\underset{x\to 1^{-}}{\lim }f\left(x\right)=-\infty \text{ and }\underset{x\to 1^{+}}{\lim }f\left(x\right)=\infty$

Explanation of Solution

Given:

The given function is $f\left(x\right)=\frac{1}{x^3-1}$ .

And the limit is $\underset{x\to 1^{-}}{\lim }\frac{1}{x^3-1}\text{ and }\underset{x\to 1^{+}}{\lim }\frac{1}{x^3-1}$

Calculation:

The given function is $f\left(x\right)=\frac{1}{x^3-1}$ .

Now draw the table for value of x that approaches from the left and from the right.

From the left, the limit is $\underset{x\to 1^{-}}{\lim }\frac{1}{x^3-1}$

x	$f\left(x\right)=\frac{1}{x^3-1}$
$0.9$	$-3.6900037$
$0.95$	$-7.011394$
$0.99$	$-33.668900$
$0.995$	$-67.001114$
$0.999$	$-333.666689$
$0.9995$	$-667.001114$
$0.9999$	$-3333.666689$
$0.99995$	$-6667.000111$
$0.99999$	$-33333.666669$

From the right, the limit is $\underset{x\to 1^{+}}{\lim }\frac{1}{x^3-1}$

x	$f\left(x\right)=\frac{1}{x^3-1}$
$1.5$	$0.421053$
$1.05$	$6.344171$
$1.01$	$33.002211$
$1.005$	$66.334442$
$1.001$	$333.000222$
$1.0005$	$666.333444$
$1.0001$	$3333.000022$
$1.00005$	$6666.333344$
$1.00001$	$33333.000002$

From the above graph, it is concluded thatthe value of given function correct to two places of decimal is $0.135$ .

b.

To determine

To find: the given limit by evaluating the given function for value of x that approaches from the left and from the right by resoning.

Answer to Problem 12E

  $\underset{x\to 1^{-}}{\lim }f\left(x\right)=-\infty \text{ and }\underset{x\to 1^{+}}{\lim }f\left(x\right)=\infty$

Explanation of Solution

Given:

The given function is $f\left(x\right)={\left(1-\frac{2}{x}\right)}^x$ .

Calculation:

The given function is $f\left(x\right)=\frac{1}{x^3-1}$ .

Now draw the table for value of x that approaches from the left and from the right.

Now draw the table of the given function $f\left(x\right)={\left(1-\frac{2}{x}\right)}^x$ and to the value of given function correct to two places of decimal.

	Numerator	Denominator	Division
$\underset{x\to 1^{-}}{\lim }\frac{1}{x^3-1}$	Positive	x close to 1and $x<1\to x^3\to x^3<0$ denominator small negative	Huge negative number; $-\infty$
$\underset{x\to 1^{+}}{\lim }\frac{1}{x^3-1}$	Positive	x close to 1and $x>1\to x^3\to x^3>0$ denominator small positive	Huge positive number; $\infty$

From the above table, it is concluded that the value of given function correct to two places of decimal is $0.135$ .

c.

To determine

To find: the given limit by evaluating the given function for value of x that approaches from the left and from the right by reasoning.

Answer to Problem 12E

  $\underset{x\to 1^{-}}{\lim }f\left(x\right)=-\infty \text{ and }\underset{x\to 1^{+}}{\lim }f\left(x\right)=\infty$

Explanation of Solution

Given:

The given function is $f\left(x\right)={\left(1-\frac{2}{x}\right)}^x$ .

Calculation:

The given function is $f\left(x\right)=\frac{1}{x^3-1}$ .

Now draw the graph of the given function:

  

There is a vertical asymptote at $x=1$

Thus, $\underset{x\to 1^{-}}{\lim }f\left(x\right)=\infty \text{ and }\underset{x\to 1^{+}}{\lim }f\left(x\right)=\infty$