Evaluate $\lim_{n \to \infty} \sqrt[n^2]{2^n+4^{n^2}}$ Evaluate $$\lim_{n \to \infty} \sqrt[n^2]{2^n+4^{n^2}}$$
We know that as $n\to \infty$ we have $2^n<<2^{2n^2}$ and therefore the limit is $4$
In a more formal way I started with:
$$\log(L)=\lim_{n \to \infty} \log(2^n+4^{n^2})^{\frac{1}{n^2}}=\lim_{n \to \infty}\frac{1}{n^2}\log(2^n+2^{2n^2})$$
Continuing to $$\log(L)=\lim_{n \to \infty}\frac{1}{n^2}\log\left[2^n(1+2^{2n})\right]$$
Did not help much
As I arrived to $$\log(L)=\lim_{n \to \infty}\frac{1}{n^2}\left[\log(2^n)+\log(1+2^{2n})\right]=\lim_{n \to \infty}\frac{1}{n^2}\log(2^n)+\lim_{n \to \infty}\frac{1}{n^2}\log(1+2^{2n})=0+\lim_{n \to \infty}\frac{1}{n^2}\log(1+2^{2n})$$

Hint:
$n>1$;
$f(n):=4(1+\dfrac{2^n}{2^{2n^2}})^{1/n^2}=$
$4(1+\dfrac{1}{2^{2n^2-n}})^{1/n^2}$;
$4(1+0)^{1/n^2} \lt f(n) < 4(1+1)^{1/n^2}.$
Take the limit.
Recall:
For $a>1$, real; and $n >1$, integer:
$1<a^{1/n^2} <a^{1/n}$, and
$\lim_{n \rightarrow \infty} a^{1/n}=1.$
