I encountered this limit and I failed, I need help with it:
$\lim_{n\to \infty } {\Huge [} {\frac{{}\sqrt[n]{2}+\sqrt[n]{3}+\sqrt[n]{5}}{3}\Huge ]}^{n} $=?
The answer they give is $\sqrt[3]{30}$
My attempt is to try solve it for case $1^{\infty }$ :
$\lim_{n\to \infty } {\Huge [} {\frac{{}\sqrt[n]{2}+\sqrt[n]{3}+\sqrt[n]{5}}{3}\Huge ]}^{n}=\lim_{n\to \infty } {\Huge [} 1+ {\frac{{}\sqrt[n]{2}+\sqrt[n]{3}+\sqrt[n]{5}-3}{3}\Huge ]}^{\frac{3}{\sqrt[n]{2}+\sqrt[n]{3}+\sqrt[n]{5}-3}\frac{\sqrt[n]{2}+\sqrt[n]{3}+\sqrt[n]{5}-3}{3}n} = e^{\lim_{n\longrightarrow \infty }\frac{n(\sqrt[n]{2}+\sqrt[n]{3}+\sqrt[n]{5}-3)}{3}}$
And now i tried to solve the :
$\lim_{n\to \infty }n(\sqrt[n]{2}+\sqrt[n]{3}+ \sqrt[n]{5}-3)$
Here i tried Stolz–Cesàro( $\frac{\infty }{\infty } $) with:
$(a_{n})_{n\ge 1} =\frac{1}{n}$
$(b_{n})_{n\ge 1} =\sqrt[n]{2}+\sqrt[n]{3}+\sqrt[n]{5}-3$
And got here
$\lim_{n\to \infty }n(n+1)(\sqrt[n]{2}+\sqrt[n]{3}+ \sqrt[n]{5}-\sqrt[n+1]{2}-\sqrt[n+1]{3}-\sqrt[n+1]{5})$
After that i was trying something with the formula : $\sqrt[n]{a}=e^{\frac{1}{n}lna}$ and got completed blocked
