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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 ?

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    $\begingroup$ That is incorrect. You can only separate limits if each exists (read: is a real number) individually. $\endgroup$ – FearfulSymmetry Jul 3 '20 at 15:14
  • $\begingroup$ 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. $\endgroup$ – Thomas Jul 3 '20 at 15:14
  • $\begingroup$ @Integrand Thanks for pointing that out. I didn't consider that. $\endgroup$ – Sibi Jul 3 '20 at 15:18
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    $\begingroup$ @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. $\endgroup$ – Peter Foreman Jul 3 '20 at 15:19
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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, $(2/3)^n\to 0$.

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

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

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    $\begingroup$ 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. $\endgroup$ – Peter Foreman Jul 3 '20 at 15:16
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Other way in general terms is: If $0<a<1$, then: Take $$y=\lim_{n\to \infty}a^n$$ $$\ln{y}=\lim_{n\to \infty}n\ln(a)$$ How $\ln(a)<0$, then $$\ln{y}\to-\infty$$ $$y\to 0$$

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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.

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