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?
 A: HINT We have
$$\dfrac12 \cdot \dfrac{3^n}{4^n} < \dfrac{2^n+3^n}{3^n+4^n} < \dfrac{3^n}{4^n}$$
A: Use the sandwich/squeeze theorem. Since
$$\frac{3^n}{2\cdot 4^n}<\frac{2^n + 3^n}{3^n + 4^n} < \frac{2\cdot 3^n}{4^n}$$
for all $n$,
$$\frac{1}{2^{1/n}} \cdot \frac{3}{4} < \sqrt[n]{\frac{2^n + 3^n}{3^n + 4^n}} < 2^{1/n}\cdot \frac{3}{4}$$
for all $n$. Since $2^{1/n} \to 1$ as $n\to \infty$, by the squeeze theorem, your limit is $3/4$.
A: The limit evaluates to... $$3 \over 4$$
Partial Proof:
$$\left({{2^n+3^n} \over {3^n+4^n}}\right)^{1/n}$$
divide the inside by $3^n$
$$\left({{(2/3)^n+1} \over {1+(4/3)^n}}\right)^{1/n}$$
$${({(2/3)^n+1)^{1/n}} \over {(1+(4/3)^n})^{1/n}}$$
The numerator evaluates to 1 and the denominator evaluates to ${4 \over 3}$
which equals ${3 \over 4}$. You'll want to verify this for yourself.
A: 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 $n$, $$\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}} \approx \sqrt[n]{\left(\frac{3}{4}\right)^n} = \frac{3}{4}$$
A: Hint:
$$
\sqrt[\large n]{\frac{2^n+3^n}{3^n+4^n}}=\frac34\ \sqrt[\large n]{\frac{1+\left(\frac23\right)^n}{1+\left(\frac34\right)^n}}
$$
