How to calculate $\lim_{n \to \infty}\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}$ I came across this strange limit whilst showing convergence of a series:
$$\lim_{n \to \infty}\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}$$
How can I calculate this limit?

Squeeze theorem gives you the proof that the limit is $\frac{3}{4}$. Since you mentioned you were looking for another way to verify that the limit is correct, here is one way (although not rigorous like the squeeze theorem) $$\begin{align}\frac{2^n+3^n}{3^n+4^n} = \frac{2^n}{3^n+4^n}+\frac{3^n}{3^n+4^n} \\ = \left(\frac{\frac{1}{2^n}}{\frac{1}{2^n}}\right)\frac{2^n}{3^n+4^n}+\left(\frac{\frac{1}{3^n}}{\frac{1}{3^n}}\right)\frac{3^n}{3^n+4^n}  \\ = \frac{1}{\left(\frac{3}{2}\right)^n+2^n}+\frac{1}{1+\left(\frac{4}{3}\right)^n}\end{align}$$ You should be able to see that $\lim_{n \to \infty} \left(\frac{3}{2}\right)^n+2^n = \infty$ so  $\lim_{n \to \infty} \frac{1}{\left(\frac{3}{2}\right)^n+2^n} = 0$. Then notice that for large $n$ the quantity $$\frac{1}{1+\left(\frac{4}{3}\right)^n} \approx \frac{1}{\left(\frac{4}{3}\right)^n} = \left(\frac{3}{4}\right)^n$$ so for large values of $n$, $$\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}} \approx \sqrt[n]{\left(\frac{3}{4}\right)^n} = \frac{3}{4}$$
