Chapter 8.2, Problem 60E

a.

To determine

To calculate: The limit by using the graphs or tables.

Answer to Problem 60E

The obtained limit is 0.

Explanation of Solution

Given Information:

The functions are defined as,

  $f\left(x\right)=\frac{1-\cos x}{x+x^2}$

Calculation:

Consider the given functions,

  $f\left(x\right)=\frac{1-\cos x}{x+x^2}$

Now, make the table for the given function.

$x$	$f\left(x\right)$
$0.1$	$\frac{1-\cos \left(0.1\right)}{0.1+{\left(0.1\right)}^2}=0.04542$
$0.01$	$\frac{1-\cos \left(0.01\right)}{0.01+{\left(0.01\right)}^2}=0.00495$
$0.001$	$\frac{1-\cos \left(0.001\right)}{0.001+{\left(0.001\right)}^2}=0.0005$
$0.0001$	$\frac{1-\cos \left(0.0001\right)}{0.0001+{\left(0.0001\right)}^2}=0.00005$

It can be seen that the limit is approaching to 0.

Therefore, the obtained limit is 0.

b.

To determine

The error in the given incorrect application of Hospital’s rule.

Answer to Problem 60E

The rule is not applicable on second line.

Explanation of Solution

Given Information:

The functions are defined as,

  $f\left(x\right)=\frac{1-\cos x}{x+x^2}$

Consider the given statements,

  $f\left(x\right)=\frac{1-\cos x}{x+x^2}$

Take the limit of the given function to 0.

  $\begin{array}{c}\underset{x\to 0}{\lim }\frac{1-\cos x}{x+x^2}=\underset{x\to 0}{\lim }\frac{0+\sin x}{1+2x}\left[\text{L'Hospitals rule}\right]\\ =\frac{\sin 0}{1+0}\\ =0\end{array}$

It can be observe that the obtained line after L’hospital the result function is not making infinite form or undefined form. So the rule is not applicable over the second line.

Therefore, the rule is not applicable on second line.