# Evaluate $\lim_{n \to \infty} \sqrt[n]{3^n+4^n}$ [duplicate]

$$\lim_{n \to \infty} \sqrt[n]{3^n+4^n}$$

Is there there a way to solve this without using $e^{ln(3^n+4^n)}$?

Maybe: $\displaystyle\lim_{n \to \infty} \sqrt[n]{4^n}=4\,\leq\,\lim_{n \to \infty} \sqrt[n]{3^n+4^n}\,\leq\,\lim_{n \to \infty} \sqrt[n]{2\cdot4^n}=4$?

## marked as duplicate by Arjang, JonMark Perry, Guy Fsone, Fabio Lucchini, muaddibJan 28 '18 at 12:32

$$\lim _{ n\rightarrow \infty }{ \sqrt [ n ]{ 3^{ n }+4^{ n } } } =4\cdot \lim _{ n\rightarrow \infty }{ \sqrt [ n ]{ \left( \frac { 3 }{ 4 } \right) ^{ n }+1 } } =4$$
Your proof is fine, provided you can use that $$\lim_{n\to\infty}\sqrt[n]{2}=1$$ This follows from Bernoulli’s inequality $(1+x)^n\ge 1+nx$, whenever $x>-1$ and $n$ is a positive integer, in the form $$\sqrt[n]{1+nx}\le 1+x$$ For $x=1/n$ this reads $$\sqrt[n]{2}\le 1+\frac{1}{n}$$ and therefore, from $$1\le\sqrt[n]{2}\le 1+\frac{1}{n}$$ and the squeeze theorem, you can conclude.
Then your application of the squeeze theorem to $$4=\sqrt[n]{4^n}\le\sqrt[n]{3^n+4^n}\le \sqrt[n]{2\cdot 4^n}=4\sqrt[n]{2}$$ is good.
$\lim_{n \to \infty} \sqrt[n]{a^n+b^n}= \text{max}(a,b)$