$\underset{x\rightarrow\infty}\lim{\Big(\sqrt{\cosh{x}}-\sqrt{\sinh{x}}\Big)}e^{\frac{3x}{2}}$ $$\underset{x\rightarrow\infty}\lim{\Big(\sqrt{\cosh{x}}-\sqrt{\sinh{x}}\Big)}e^{\frac{3x}{2}}$$
$$\underset{x\rightarrow\infty}\lim{\Bigg(\sqrt{\frac{e^{4x}+e^{2x}}{2}}-\sqrt{\frac{e^{4x}-e^{2x}}{2}}\Bigg)}$$
Should I use L'Hopital's s rule from this point?
 A: I think a natural approach would be\begin{align}\lim_{x\to\infty}\left(\sqrt{\cosh x}-\sqrt{\sinh x}\right)e^{3x/2}&=\lim_{x\to\infty}\frac{\left(\sqrt{\cosh x}-\sqrt{\sinh x}\right)\left(\sqrt{\cosh x}+\sqrt{\sinh x}\right)}{\sqrt{\cosh x}+\sqrt{\sinh x}}e^{3x/2}\\&=\lim_{x\to\infty}\frac{e^{x/2}}{\sqrt{\frac{e^x+e^{-x}}2}+\sqrt{\frac{e^x-e^{-x}}2}}.\end{align}Can you take it from here?
A: For those, Taylor expansions are very useful and give a systematic way to solve the problem. You have
$$\begin{align}
\Big(\sqrt{\cosh{x}}-\sqrt{\sinh{x}}\Big)e^{\frac{3x}{2}}
&= \Big(\sqrt{\frac{e^x+e^{-x}}{2}}-\sqrt{\frac{e^x-e^{-x}}{2}}\Big)e^{\frac{3x}{2}} \\
&=\frac{e^{\frac{x}{2}}}{\sqrt{2}}\Big(\sqrt{1+e^{-2x}}-\sqrt{1-e^{-2x}}\Big)e^{\frac{3x}{2}} \\
&=\frac{e^{2x}}{\sqrt{2}}\Big(\Big(1+\frac{e^{-2x}}{2}+ o(e^{-2x})\Big)-\Big(1-\frac{e^{-2x}}{2}+ o(e^{-2x})\Big)\Big) \\
&=\frac{e^{2x}}{\sqrt{2}}\Big(e^{-2x} + o(e^{-2x})\Big) \\
&=\frac{1}{\sqrt{2}}(1 + o(1)) \xrightarrow[x\to\infty]{} \boxed{\frac{1}{\sqrt{2}}}
\end{align}$$
where we used the fact that $\sqrt{1+u} = 1+\frac{u}{2} + o(u)$ when $u\to 0$.
