Limit Exercise asked in the brilliant.org

Exercise:

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

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