What is the solution of $\lim_{n\rightarrow +\infty}(2+3^n)^{\frac{1}{2n}}$? Here's what I did: $$\lim_{n\rightarrow +\infty}(2+3^n)^{\frac{1}{2n}}=\lim_{n\rightarrow +\infty}e^{\frac{1}{2n}\ln(2+3^n)}$$
What should I do next in order to solve it?
Note that
$$
3^n \le 2+3^n\le 2\cdot3^n
$$
Hence
$$
3^{1/2} \le (2+3^n)^{1/2n}\le 2^{1/2n}\cdot3^{1/2}\to 3^{1/2}.
$$
$3^{(1/2)}((2/3^n)+1)^{(1/(2n))}$.

For $n$ large enough: $2/(3^n) &lt;1$:

$\small {  √3 \lt √3(2/3^n+1)^{(1/(2n))} \lt √3(√2)^{(1/n)}}$.

Squeeze.