Question

$\displaystyle \lim _{x \rightarrow \infty}\left(\frac{x^{100}}{e^{x}}+\left(\cos \frac{2}{x}\right)^{x^{2}}\right)=$

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

Answer 1

Answer: (D)

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