Question

$\displaystyle \lim_{x\to\infty}\left(\frac{x^{100}}{e^{x}}+\left(cos \frac{2}{x}\right)^{x^2}\right) = $

• A. $e^{-1}$
• B. $e^{-4}$
• C. $(1+e^{-2})$
• D. $e^{-2}$

Answer 1

Answer: (D)

Consider $\displaystyle \lim_{x\to\infty} \left[\frac{x^{100} }{e^{x}}+\left(cos \frac{2}{x}\right)^{x^2}\right] $
$ = \displaystyle \lim _{x\to \infty } \frac{x^{100} }{e^{x}}+\lim _{x\to \infty }\left[\left(cos \frac{2}{x}\right)\right]^{x^2}$
$= \displaystyle \lim _{x\to \infty } \frac{x^{100} }{e^{x}} = 0$ (Using L' Hopital’s rule)
and $\displaystyle\lim _{x\to \infty }\left(cos \frac{2}{x}\right)^{x^2}$ is of $\left(1^{\infty}\right)$ form
$= e^{\displaystyle \lim_{x \to\infty}} x^{2}\left(cos \frac{2}{x}-1\right)$
$= e^{\displaystyle\lim_{t \to 0}} t^{\frac{4}{2}\left(cost-1\right)}$ (Put $\frac{2}{x} = t \Rightarrow x= \frac{2}{t}$)
$= e^{\displaystyle -\lim_{t \to 0}} \left(\frac{1-cost}{t^{2}}\right)4 = e^{\displaystyle-\lim_{t \to 0}}\left(\frac{sint}{2t}\right)4 = e^{-2}$