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.
 A: Let the limit exist and equal $L$. Then $\ln L$ equals:
$$\lim_{x \to \infty} x\left(\ln\left(2^{1/x}+3^{1/x}\right)-\ln\left(4^{1/x}+5^{1/x}\right)\right)$$
and making the substitution $u \mapsto 1/x$ gives:
$$\lim_{u \to 0^+} \frac{1}{u} \left(\ln\left(2^u+3^u\right)-\ln\left(4^u+5^u\right)\right)$$
Considering the first-order approximation of $f(u) = \ln\left(2^u+3^u\right)$ using Taylor series, $f(0) = \ln 2$. $f'(u) = \frac{1}{2^u + 3^u} \cdot (2^u \ln 2 + 2^u \ln 3)$, hence $f'(0) = \frac{\ln 6}{2}$. Doing this for the other composite function gives:
$$\ln L = \lim_{u \to 0^+} \frac{1}{u} \left(\ln 2 + \frac{\ln 6}{2}u - \ln 2 - \frac{\ln 20}{2} u \right) = \frac{1}{2} \ln \frac{3}{10}$$
hence $L = e^{\ln(3/10) \cdot 1/2} = \sqrt{3/10} \approx 0.548$.
A: 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.
