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. $(1+e^{-2})$
• D. $e^{-2}$

Answer 1

Answer: (D)

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