Question

What is the limit as x approaches infinity of $x\tan \left( \dfrac{1}{x} \right)$?

Answer 1

Assume the given limit as L. Now, replace x with $\dfrac{1}{h}$ and find the values that h tends to using the relation $x \to \infty$. Use the fact that any non – zero number divided by infinity is 0. Simplify the limit into the form $\displaystyle \lim_{h\to 0}\dfrac{\tan h}{h}$ and use the basic formula $\displaystyle \lim_{\theta \to 0}\dfrac{\tan \theta }{\theta }=1$ to get the answer.

Complete step-by-step solution:
Here we have been provided with the expression $x\tan \left( \dfrac{1}{x} \right)$ and we are asked to find its limit value as x is tending to infinity. Let us assume this limit as L, so we have,
$\Rightarrow L=\displaystyle \lim_{x \to \infty }\left[ x\tan \left( \dfrac{1}{x} \right) \right]$
Here we can see that we cannot directly substitute the value of x to get the answer because if we do so then the limit will become $\infty \times \tan \left( \infty \right)$ which is an indeterminate form. So we need a different approach. Now, replacing x with $\dfrac{1}{h}$ we get,
$\Rightarrow \dfrac{1}{h}\to \infty$
Taking reciprocal both the sides we get,
$\Rightarrow h\to \dfrac{1}{\infty }$
We know that any non – zero real number divided by infinity is 0 so we get,
$\Rightarrow h\to 0$
Substituting these values in the expression L we get,
$\begin{aligned} & \Rightarrow L=\displaystyle \lim_{h\to 0}\left[ \dfrac{1}{h}\tan \left( h \right) \right] \\\ &\Rightarrow L=\displaystyle \lim_{h\to 0}\left[ \dfrac{\tan \left( h \right)}{h} \right] \\\ \end{aligned}$
Using the basic formula of limit given as $\displaystyle \lim_{\theta \to 0}\dfrac{\tan \theta }{\theta }=1$ we get,
$\therefore L=1$
Hence, the value of the given expression of limit is 1.

Note: There are two other methods by which you can solve the above question. The easiest one is called the L Hospital’s rule. This method is used when we have the limit of the form $\displaystyle \lim_{x \to0}\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{0}{0}$ or $\displaystyle \lim_{x \to 0}\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{\infty }{\infty }$. Here what we do is, we differentiate $f\left( x \right)$ and $g\left( x \right)$ to get the limit of the form $\displaystyle \lim_{x \to 0}\dfrac{f'\left( x\right)}{g'\left( x \right)}$ and check if it is still in indeterminate form or not. If it is not then we substitute the value of x to get the answer and if it is still in indeterminate form then we again differentiate and this process continues till it reaches its determinate form. The other method is the expansion formula of $\tan x$ given as $\tan x=x+\dfrac{{{x}^{3}}}{3}+\dfrac{2{{x}^{5}}}{15}+......$.
Substitute this value in the limit, cancel the like terms and at last substitute the value of x to get the answer. Remember that the two processes that we have mentioned here in the note must be applied only after we replace x with $\dfrac{1}{h}$.