How to calculate $\lim_{n \to \infty } \left( {\sqrt[n]{{{3^n} + {4^n}}}} \right)$? How to calculate $$\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt[n]{{{3^n} + {4^n}}}} \right)?$$

I saw on the suggestions that you can use the sandwich theorem or compression. And the fact that $\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt[n]{2}} \right) = 1$
So I must first find upper bounds and lower.
someone help me with this please?
 A: Complete Solution: Sandwich theorem
$$ \sqrt[n]{4^n} \le \sqrt[n]{4^n + 3^n} \le \sqrt[n]{2\times4^n}$$
and so
$$ 4 \le \sqrt[n]{4^n + 3^n}  \le 2^{1/n} \times 4 $$
Since $2^{1/n} \to 1$ as $n \to \infty$, Hence the required limit is $4$.
OR, use a sledgehammer:
Use the fact that for positive $a_n$ if $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$ exists then so does $\lim\limits_{n\rightarrow\infty}\root n\of {a_n}$ and they are equal.Here $a_n=4^n+3^n$ and one can show
$$
\lim\limits_{n\rightarrow\infty} {4^{n+1}+3^{n+1}\over 4^n+3^n} =4.
$$
So, then  $\lim\limits_{n\rightarrow\infty} \root n\of{4^n+3^n}=4$ as well.
A: Hint.
$$\sqrt[n]{{{3^n} + {4^n}}}=\sqrt[n]{4^n(1+\left(\frac{3}{4}\right)^n)}=4\sqrt[n]{1+\left(\frac{3}{4}\right)^n}$$
A: $$3^n + 4^n \sim 4^n \implies \sqrt[n] {3^n + 4^n} \sim \sqrt[n]{4^n} = 4$$
A: You can also do the following:
$$ (3^n+4^n)^{\frac{1}{n}}=e^{\frac{1}{n}\ln(3^n+4^n)}=e^{\frac{1}{n}\ln[4^n(1+\frac{3^n}{4^n})]}$$
$$ = e^{\frac{1}{n}[n\ln 4 + \ln (1+\frac{3^n}{4^n})]}=e^{\ln 4 + \frac{\ln (1+\frac{3^n}{4^n})}{n}}=4\cdot e^{\frac{\ln (1+\frac{3^n}{4^n})}{n}}$$
It is now easy to show that $e^{\frac{\ln (1+\frac{3^n}{4^n})}{n}}$ goes to $0$ for $ n \rightarrow \infty $.
