evaluate $\lim_{n \rightarrow \infty} \sqrt[n]{2^n+3^n}$

Question

I am interested in evaluating the following limit: $$\lim_{n \rightarrow \infty} \sqrt[n]{2^n+3^n}$$ I realize it should be 3 because as $n$ approaches infinity, $3^n$ gets significantly larger than $2^n$.
But numerically: $$\lim_{n \rightarrow \infty} \sqrt[n]{2^n+3^n}= \lim_{n \rightarrow \infty} \sqrt[n]{\frac{2^n+3^n}{1}}= \lim_{n \rightarrow \infty} \sqrt[n]{\frac{(\frac{2}{3})^n+1}{\frac{1}{3^n}}}$$ which is $\sqrt[\infty]{\frac{2}{0}}$ ..?
Where is my mistake?

Answer 1

Hint

Note that $$3=\sqrt[n]{3^n}<\sqrt[n]{2^n+3^n}<\sqrt[n]{2\cdot 3^n}=3\cdot 2^{1\over n}$$

Answer 2

Try $$\sqrt[n]{2^n+3^n}=\sqrt[n]{3^n\left(1+(\tfrac23)^n\right)} =3\cdot \sqrt[n]{1+(\tfrac23)^n}$$ instead.