Calculate $\lim_{x\to \infty}(\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}})^x$ Calculate $\lim_{x\to \infty}(\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}})^x$.
First off, it's easy to see that $\lim_{x\to \infty}\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}}$ = 1. Therefore, I tried the following:
$$\lim_{x\to \infty}(\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}})^x=
\lim_{x\to \infty}(1 +(\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}} -1))^x = e^{\lim_{x\to \infty}x(\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}} -1)}.$$
Now I find myself stuck at finding $\lim_{x\to \infty}x(\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}} -1)$. Keep in mind that I am not allowed to use l'Hopital. Any hint would be appreciated. Thanks.

In fact $\lim_{x\to\infty}\left(\frac{\sqrt[x]{a}+\sqrt[x]{b}}{2}\right)^x=\sqrt{ab}$ for $a,\,b>0$. This is a comparison (in the $2$-variable unweighted case) of power means to the geometric mean. Now just take the ratio of two cases.
