$\underset{x\rightarrow\infty}\lim{\Big(\sqrt{\cosh{x}}-\sqrt{\sinh{x}}\Big)}e^{\frac{3x}{2}}$

Question

$$\underset{x\rightarrow\infty}\lim{\Big(\sqrt{\cosh{x}}-\sqrt{\sinh{x}}\Big)}e^{\frac{3x}{2}}$$ $$\underset{x\rightarrow\infty}\lim{\Bigg(\sqrt{\frac{e^{4x}+e^{2x}}{2}}-\sqrt{\frac{e^{4x}-e^{2x}}{2}}\Bigg)}$$

Should I use L'Hopital's s rule from this point?

Answer 1

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 $u\to 0$.

Answer 2

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?