2011/043) find limit $\lim_{n\to\infty} \sqrt[n]{3^n+4^n}$

We can we take 3 or 4 out but taking 3 it diverges

we get $3 \sqrt[n]{1^n+(\frac{4}{3})^n}$ and reach no where

taking 4 we get
$4 \sqrt[n]{1^n+(\frac{3}{4})^n}$

now $1 < 1+(\frac{3}{4})^n < 2$

so  $\sqrt[n]{1} < \sqrt[n]{1+(\frac{3}{4})^n} < \sqrt[n]{2} $ and as $\sqrt[n]{2}$ as n goes to infinite goes to 1

so result = 4 * 1 or 4.