Cool Limit.

First we use conjugates to find $\lim_{x\to\infty}(\sqrt{x^4+1}-(x^2+1))$ = $\lim_{x\to\infty}\frac{(\sqrt{x^4+1}-(x^2+1))(\sqrt{x^4+1}+(x^2+1))}{\sqrt{x^4+1}+(x^2+1)}$ = $\lim_{x\to\infty}\frac{-2x^2}{\sqrt{x^4+1}+(x^2+1)}$ = $\lim_{x\to\infty}\frac{-2}{\sqrt{1+1/x^4}+(1+1/x^2)}$ = $-1$ . Now $\lim_{x\to\infty}(e^{\sqrt{x^4+1}}-e^{x^2+1})$ = $\lim_{x\to\infty}e^{x^2+1}(e^{\sqrt{x^4+1}-(x^2+1)}-1)$ = $-\infty$ since the first factor, $e^{x^2+1}$ , goes to infinity and the second factor goes to $e^{-1}-1<0$ .

Solution: we have ${{e}^{\sqrt{{{x}^{4}}+1}}}-{{e}^{{{x}^{2}}+1}}={{e}^{\sqrt{{{x}^{4}}+1}}}\left( 1-\frac{{{e}^{{{x}^{2}}+1}}}{{{e}^{\sqrt{{{x}^{4}}+1}}}} \right)={{e}^{\sqrt{{{x}^{4}}+1}}}\left( 1-{{e}^{{{x}^{2}}+1-\sqrt{{{x}^{4}}+1}}} \right)$

So $\underset{x\to \infty }{\mathop{\lim }}\,\left( {{e}^{\sqrt{{{x}^{4}}+1}}}-{{e}^{{{x}^{2}}+1}} \right)=\underset{x\to \infty }{\mathop{\lim }}\,{{e}^{\sqrt{{{x}^{4}}+1}}}\left( 1-\underset{x\to \infty }{\mathop{\lim }}\,{{e}^{{{x}^{2}}+1-\sqrt{{{x}^{4}}+1}}} \right)$ but $e\$$ is continuous function

So $\underset{x\to \infty }{\mathop{\lim }}\,{{e}^{{{x}^{2}}+1-\sqrt{{{x}^{4}}+1}}}={{e}^{\underset{x\to \infty }{\mathop{\lim }}\,\left( {{x}^{2}}+1-\sqrt{{{x}^{4}}+1} \right)}}$ but $\underset{x\to \infty }{\mathop{\lim }}\,\left( {{x}^{2}}+1-\sqrt{{{x}^{4}}+1} \right)=1$

So $\underset{x\to \infty }{\mathop{\lim }}\,\left( {{e}^{\sqrt{{{x}^{4}}+1}}}-{{e}^{{{x}^{2}}+1}} \right)={{e}^{\infty }}\left( \underbrace{1-e}_{\le 0} \right)=-\infty$

Note that , $\underset{x\to \infty }{\mathop{\lim }}\,\left( {{x}^{2}}+1-\sqrt{{{x}^{4}}+1} \right)=\underset{x\to \infty }{\mathop{\lim }}\,\left( {{x}^{2}}+1-\sqrt{{{x}^{4}}\left( 1+\frac{1}{{{x}^{4}}} \right)} \right)$

$=\underset{x\to \infty }{\mathop{\lim }}\,\left( {{x}^{2}}\left( 1-\sqrt{1+\frac{1}{{{x}^{4}}}} \right)+1 \right)=1+\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\left( 1-\sqrt{1+\frac{1}{{{x}^{4}}}} \right)=1+\infty \left( 0 \right)$

But $\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\left( 1-\sqrt{1+\frac{1}{{{x}^{4}}}} \right)=\underset{x\to \infty }{\mathop{\lim }}\,\frac{1-\sqrt{1+\frac{1}{{{x}^{4}}}}}{\frac{1}{{{x}^{2}}}}$

Take $u=\frac{1}{{{x}^{2}}}\Rightarrow {{u}^{2}}=\frac{1}{{{x}^{4}}}$ as $x\xrightarrow{{}}\infty$ , $u\xrightarrow{{}}0$

So $\underset{u\to 0}{\mathop{\lim }}\,\frac{1-\sqrt{1+{{u}^{2}}}}{u}\times \frac{1+\sqrt{1+{{u}^{2}}}}{1+\sqrt{1+{{u}^{2}}}}=\underset{u\to 0}{\mathop{\lim }}\,\frac{1-\left( 1+{{u}^{2}} \right)}{u\left( 1+\sqrt{1+{{u}^{2}}} \right)}=\underset{u\to 0}{\mathop{\lim }}\,\frac{-{{u}^{2}}}{u\left( 1+\sqrt{1+{{u}^{2}}} \right)}$

$=\underset{u\to 0}{\mathop{\lim }}\,\frac{-u}{1+\sqrt{1+{{u}^{2}}}}=\frac{0}{1+1}=\frac{0}{2}=0$ Thus $\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\left( 1-\sqrt{1+\frac{1}{{{x}^{4}}}} \right)=0$

The given expression is :- $\lim_{x\to\infty} \Bigg( e^{\sqrt{x^{4}+1}}-e^{x^{2}+1}\Bigg)$ We can write it in series form as follows:- $\lim_{x\to\infty} \sum_{r=0}^{\infty} \Bigg(\dfrac{(x^{4}+1)^{\frac{r}{2}}}{r!} - \dfrac{(x^{2}+1)^{r}}{r!}\Bigg)$ $\lim_{x\to\infty} \sum_{r=0}^{\infty} \dfrac{(x^{4}+1)^{\frac{r}{2}}-(x^{4}+1+2x^{2})^{\frac{r}{2}}}{r!}$ Here we see that in every term some negative quantity is added.

Hence when all such terms are added a large negative quantity is left which is $-\infty$ .

The inequality $\sqrt{x^{4}+1} \leq x^{2}+1$ is easy to prove.

Since $\frac{d}{dx} e^{x} = e^{x}$ and $e^{x} > 0$ for all $x \in \mathbb{R}$ , $e^{x}$ is hence a strictly monotonically increasing function. Thus, $e^{\sqrt{x^{4}+1}}-e^{x^{2}+1}$ is negative and unbounded, ie $\rightarrow -\infty$ .