# Alternative way to solve this limit?

The solution to this limit should be 3. I know that it can be solved by using the squeeze theorem, by coming up with two other sequences whose limit is 3, but I would prefer some other method if possible as I'm not comfortable with this one. Is there any other way to solve it?

$\lim _{n\to \infty \:}\left(\left(2^n+3^n\right)^{\frac{1}{n}}\right)$

Thank you.

We have

$$2^n=_\infty o(3^n)$$ so $$(2^n+3^n)^{\frac1n}\sim_\infty (3^n)^{\frac1n}=3$$

I know that while the function $\alpha\approx0$ then $\sqrt[n]{1+\alpha}\approx\frac{\alpha}{n}+1$. Now a suitable factor inside the brackets can lead us to $3$.

Sooner or later, you probably have to reduce it to the squeeze theorem. But here is a solution which hides it a little bit: $$(2^n+3^n)^{1/n}=3(1+(2/3)^n)^{1/n},$$ so we only have to show that $(1+(2/3)^n)^{1/n}\rightarrow 1$. Consider the fact that $$(1+(2/3)^n)^{1/n}=\exp\left(\frac{1}{n}\ln (1+(2/3)^n)\right).$$ By continuity of the logarithm, $\ln(1+(2/3)^n)\rightarrow 0$, since $(2/3)^n\rightarrow 0$. It follows that $\ln (1+(2/3)^n)/n\rightarrow 0$. By continuity of the exponential function, this implies $(1+(2/3)^n)^{1/n}\rightarrow 1$.

But it would have been much simpler to simply assert that $$1\leq (1+(2/3)^n)^{1/n}\leq 2^{1/n}.$$

Thank you all for responding, but I like this answer (that I found in a different question on this website) the best:

\begin{align} \lim_{n\to\infty}\left(3^n+5^n\right)^{1/n} &=5\lim_{n\to\infty}\left(\left(\frac35\right)^n+1\right)^{1/n}\\ &=5(0+1)^0\\[9pt] &=5 \end{align}

The numbers are different, but it's the same otherwise.