Evaluate $\lim_{n \to \infty} \sqrt[n^2]{2^n+4^{n^2}}$

Question

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 $$\log(L)=\lim_{n \to \infty}\frac{1}{n^2}\left[\log(2^n)+\log(1+2^{2n})\right]=\lim_{n \to \infty}\frac{1}{n^2}\log(2^n)+\lim_{n \to \infty}\frac{1}{n^2}\log(1+2^{2n})=0+\lim_{n \to \infty}\frac{1}{n^2}\log(1+2^{2n})$$

Answer 1

Hint: $$\frac{\ln\left(0+4^{x^2}\right)}{x^2}\le\frac{\ln\left(2^{x}+4^{x^{2}}\right)}{x^{2}}\le\frac{\ln\left(4^{x^{2}}+4^{x^{2}}\right)}{x^{2}} $$

Answer 2

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

$\lim_{n \rightarrow \infty} a^{1/n}=1.$