Question

Let $f(x)=\frac{\ln \left(x^{2}+e^{x}\right)}{\ln \left(x^{4}+e^{2 x}\right)}$, then value of $\underset{x \rightarrow \infty}{\text{Lim}}\left(\frac{1}{f(x)}\right)^{f(x)}$ is equal to

Answer 1

$\underset{x \rightarrow \infty}{\text{Lim}} f(x)=\underset{x \rightarrow \infty}{\text{Lim}}\frac{x+\ln \left(e^{-x} x^{2}+1\right)}{2 x+\ln \left(e^{-x} x^{4}+1\right)}=\frac{1}{2}$
$\because \underset{x \rightarrow \infty}{\text{Lim}}e^{-x} x^{4}=\underset{x \rightarrow \infty}{\text{Lim}} \frac{x^{4}}{e^{x}}=0$
$\therefore =2^{\frac{1}{2}}=\sqrt{2}$