Question

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

• A. $e^2$
• B. $e^{-2}$
• C. $e^4$
• D. $e^{-4}$

Answer 1

The correct option is B $e^{-2}$

$\underset{x\to \infty }{lim}\{\frac{x^{100}}{e^x}+{\left(\cos \frac{2}{x}\right)}^{x^2}\}=\underset{x\to \infty }{lim}\frac{x^{100}}{e^x}+\underset{x\to \infty }{lim}{\left(\cos \frac{2}{x}\right)}^{x^2}\Downarrow 0=\underset{x\to \infty }{lim}{\left(\cos \frac{2}{x}\right)}^{x^2}\left(1^{\infty }\text{form}\right)=e^{\underset{x\to \infty }{lim}{x}^2(\cos \frac{2}{x}-1)}$
Let $\frac{2}{x}=t\therefore e^{\underset{x\to \infty }{lim}{x}^2(\cos \frac{2}{x}-1)}=e^{\underset{t\to 0}{lim}4\frac{(\cos t-1)}{t^2}}=e^{-4\times \frac{1}{2}}=e^{-2}$