Chapter 4.5, Problem 32E

To determine

The value of given limit.

Answer to Problem 32E

  $1$ .

Explanation of Solution

Given information:

A limit function is given as,

  $\underset{x\to \infty }{\lim }x\tan \left(\frac{1}{x}\right)$ .

Concept used:

L'Hospital's Rule: If $f$ and $g$ are differentiable in an open interval containing $a$ and $f(a)=g(a)=0$ . Also, $g\text{'}(x)\ne 0$ (except possibly at $a$ ). Then-

  $\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f\text{'}(x)}{g\text{'}(x)}$

This rule is also applicable when $f(a)=\pm \infty \text{ and }g(a)=\pm \infty$ .

Calculation:

Given limit function is, $\underset{x\to \infty }{\lim }x\tan \left(\frac{1}{x}\right)$ .

Here limit converts into an indeterminate form of $\infty \cdot 0$ .

Rewrite the limit function,

  $\begin{array}{l}=\underset{x\to \infty }{\lim }\frac{x\sin \left(\frac{1}{x}\right)}{\cos \left(\frac{1}{x}\right)}\\ =\frac{\underset{x\to \infty }{\lim }x\sin \left(\frac{1}{x}\right)}{\underset{x\to \infty }{\lim }\cos \left(\frac{1}{x}\right)}\end{array}$

L'Hospital's Rule is applicable here.

Using L'Hospital's Rule,

  $\begin{array}{l}\text{Let us first calculate limit function in nemerator }\\ =\underset{x\to \infty }{\lim }x\sin \left(\frac{1}{x}\right)\\ =\underset{x\to \infty }{\lim }\frac{\sin \left(\frac{1}{x}\right)}{\frac{1}{x}}\\ \text{Apply L'Hospital's rule,}\\ =\underset{x\to \infty }{\lim }\frac{-\frac{\cos \left(\frac{1}{x}\right)}{x^2}}{-\frac{1}{x^2}}\\ =\underset{x\to \infty }{\lim }\cos \left(\frac{1}{x}\right)\\ \text{Now given limit function is, }\\ \frac{\underset{x\to \infty }{\lim }x\sin \left(\frac{1}{x}\right)}{\underset{x\to \infty }{\lim }\cos \left(\frac{1}{x}\right)}\\ =\frac{\underset{x\to \infty }{\lim }\cos \left(\frac{1}{x}\right)}{\underset{x\to \infty }{\lim }\cos \left(\frac{1}{x}\right)}\\ \text{Plug in the value }x=\infty ,\\ =\frac{\cos 0}{\cos 0}\\ =1\end{array}$

Hence, value of given limit function is $1$ .