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I am trying to determine if 3$^n$ grows faster than 2$^{2n}$.

One way I found online to do this was, from Growth was to evaluate $\lim_{n\to \infty} \frac{3^n}{2^{2n}}$ and if that limit evaluates to infinity, then 3$^n$ grows faster than 2$^{2n}$.

I am trouble with evaluating this limit though. In my initial evaluation of this limit, I saw that say you plugged in a really large $n$, you get the indeterminate form $ \frac{\infty}{\infty}$ so from here L Hopital, you can use L’Hospital’s Rule.

However when I tried to use L'Hospital's Rule once, this is what I got
    $\lim_{n\to \infty} \frac{3^nln(3)}{2^{2n}2ln(2)}$

I think this is mathematically correct but I couldn't find a way to reduce this further. I also tried an approach from Exponent L Hospital but I couldn't take the natural log of both sides because the two exponents are of different bases.

Am I going about this the right way? Is there another way of evaluating this limit?

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    $\begingroup$ Note that $2^{2n}=4^n$. $\endgroup$ – Gregory Grant Apr 6 '15 at 19:07
  • $\begingroup$ $\frac{3^{n}}{2^{2n}}=\frac{3^{n}}{4^{n}}=(\frac{3}{4})^{n}$. Now you can use the fact that $0<\frac{3}{4}<1$. $\endgroup$ – mich95 Apr 6 '15 at 19:09
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Inddeed you have: $$\lim_{n\to \infty} \frac{3^n}{2^{2n}}=\lim_{n\to \infty} \frac{3^n}{4^{n}}=\lim_{n\to \infty} \left(\frac{3}{4}\right)^n=0$$

because $\frac{3}{4}<1$

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  • $\begingroup$ What exponent rule did you use to simplify $2^{2n}$ to $4^n$? $\endgroup$ – committedandroider Apr 6 '15 at 19:24
  • $\begingroup$ I used $x^{ab}=(x^a)^b$ for $x=2, a=2, b=n$ $\endgroup$ – Elaqqad Apr 6 '15 at 19:25
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Since $2^{2n}=4^n$, you have $\frac{3^n}{2^{2n}}=\left(\frac34\right)^n\rightarrow 0$.

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  • $\begingroup$ What exponent rule did you use to simplify $2^{2n}$ to $4^n$? $\endgroup$ – committedandroider Apr 6 '15 at 19:24
  • $\begingroup$ $(a^b)^c=a^{bc}$. $\endgroup$ – Gregory Grant Apr 6 '15 at 19:27
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There are multiple ways to complete this problem as mentioned by other answers. You mentioned you didn't know how to do the logarithmic way. This is how:

$$\begin{align}L & = \lim_{n\to\infty}\frac{3^n}{2^{2n}} \\ \ln(L) & = \lim_{n\to\infty}\ln\left(\frac{3^n}{2^{2n}}\right) \\ \ln(L) & = \lim_{n\to\infty} \left[\ln(3^n)-\ln(2^{2n})\right] \\ \ln(L) & = \lim_{n\to\infty}\left[n\ln(3) - n\ln(2^2)\right]\\ \ln(L) &=\lim_{n\to\infty}n\ln\left(\frac{3}{4}\right)\\ \ln(L) & = \ln\left(\frac{3}{4}\right)\lim_{n\to\infty}n\end{align}$$

$\ln(3/4) < 0$, so it produces a negative value. Thus,

$$\begin{align}\ln(L) & \to -\lim_{n\to\infty}n \\e^{\ln(L)} & = e^{ -\lim_{n\to\infty}n} \\ L & = \frac{1}{e^{\lim_{n\to\infty}}}\\L & = 0\end{align}$$

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