# Evaluate $\lim_{n \to \infty} \frac{2^n}{3^n}$

As stated in the question, I'm trying to find the limit $$\lim_{n \to \infty} \frac{2^n}{3^n}$$

This is my attempt:

$$\lim_{n \to \infty} \frac{2^n}{3^n} = \lim_{n \to \infty} 2^n \cdot \lim_{n \to \infty} \frac{1}{3^n}$$

The first limit pulls to $$\infty$$ whereas the second limit pulls to $$0$$ and hence the limit will be $$0$$. Is the justfication right ?

Is there any other way to solve it ?

• That is incorrect. You can only separate limits if each exists (read: is a real number) individually. – FearfulSymmetry Jul 3 '20 at 15:14
• While your conclusion is correct the reasoning is not. Pulling apart the expression in two limits is not a good idea and the $=$ sign you've written down is actually incorrect. – Thomas Jul 3 '20 at 15:14
• @Integrand Thanks for pointing that out. I didn't consider that. – Sibi Jul 3 '20 at 15:18
• @Sibi Even if you could seperate the limits it doesn't make any sense to say $\infty\cdot0=0$ because this is a well known indeterminate form. – Peter Foreman Jul 3 '20 at 15:19

Claim: for $$n\geq 1$$, $$n/2<(3/2)^n$$. The claim is immediate for $$n=1$$ and follows by induction: the LHS increases by $$1/2$$ as $$n$$ increases to $$n+1$$ and the RHS increases by at least $$3/4$$ (by much more, in fact, but this is sufficient).

Then $$0 <(2/3)^n<2/n$$, and $$2/n\to 0$$. By the Squeeze Theorem, $$(2/3)^n\to 0$$.

• Noce answer. The claim is true for any $n \in \Bbb {N}$ I have already upvoted your answer. – Aryadeva Sep 25 '20 at 19:19

Let $$r\in(0,1)$$. Then the sequence $$a_{n} = r^{n}$$ is bounded below by zero and it is decreasing. Thus it converges.

Moreover, we have that \begin{align*} L = \lim_{n\to\infty}r^{n+1} = \lim_{n\to\infty}r\times r^{n} = r\lim_{n\to\infty}r^{n} = rL \Rightarrow L(1 - r) = 0 \end{align*} Since $$r\in(0,1)$$, we conclude that $$L = 0$$. At your case, $$r = 2/3$$.

Hopefully this helps.

You can write $$\frac{2^n}{3^n}$$ as $$\left(\frac{2}{3}\right)^n$$ and say that the limit is 0 as $$\frac{2}{3}<1$$.

• I don't like this answer because you seem to be applying the fact that $|a|\lt1\implies \lim_{n\to\infty}a^n=0$ which is much stronger than the OPs question. – Peter Foreman Jul 3 '20 at 15:16

Other way in general terms is: If $$0, then: Take $$y=\lim_{n\to \infty}a^n$$ $$\ln{y}=\lim_{n\to \infty}n\ln(a)$$ How $$\ln(a)<0$$, then $$\ln{y}\to-\infty$$ $$y\to 0$$

An intuitive approach would be to observe that $$3^n/2^n = (1.5)^n$$ tends to $$\infty$$ as $$n$$ goes to infinity.

So $$1/(1.5)^n$$ goes to $$0$$ as $$n$$ goes to infinity.