Question

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

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

Answer 1

Answer: (D)

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