1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

We have

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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 $3$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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 $$ 1\leq (1+(2/3)^n)^{1/n}\leq 2^{1/n}. $$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.