#### The value of $\lim _{x \rightarrow \infty}\left(e^{\sqrt{x^4+1}}-e^{\left(x^2+1\right)}\right)$ is Option: 1 $0$Option: 2 $e$Option: 3 $1/e$Option: 4 $-\infty$

$L=\lim _{x \rightarrow \infty}\left(e^{\sqrt{x^4+1}}-e^{\left(x^2+1\right)}\right)$

$=\lim _{x \rightarrow \infty} e^{x^2+1}\left[e^{\sqrt{x^4+1}-\left(x^2+1\right)}-1\right]$

Now,

\begin{aligned} & \lim _{x \rightarrow \infty}\left[\sqrt{x^4+1}-\left(x^2+1\right)\right] \\ \\& =\lim _{x \rightarrow \infty} \frac{x^4+1-\left(x^4+1+2 x^2\right)}{\sqrt{x^4+1}+\left(x^2+1\right)} \end{aligned}

\begin{aligned} & =\lim _{x \rightarrow \infty} \frac{-2 x^2}{x^2\left(\sqrt{1+\left(1 / x^4\right)}+1+\left(1 / x^2\right)\right)} \\ \\& =-1 \\ \\& \therefore \quad L=\infty \times\left(\frac{1}{e}-1\right) \\ \\& =-\infty \end{aligned}\begin{aligned} & =\lim _{x \rightarrow \infty} \frac{-2 x^2}{x^2\left(\sqrt{1+\left(1 / x^4\right)}+1+\left(1 / x^2\right)\right)} \\ \\& =-1 \\ \\& \therefore \quad L=\infty \times\left(\frac{1}{e}-1\right) \\ \\& =-\infty \end{aligned}