# How to evaluate this exponential fraction limit?

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.

• Note that $2^{2n}=4^n$. Apr 6, 2015 at 19:07
• $\frac{3^{n}}{2^{2n}}=\frac{3^{n}}{4^{n}}=(\frac{3}{4})^{n}$. Now you can use the fact that $0<\frac{3}{4}<1$. Apr 6, 2015 at 19:09

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

• What exponent rule did you use to simplify $2^{2n}$ to $4^n$? Apr 6, 2015 at 19:24
• I used $x^{ab}=(x^a)^b$ for $x=2, a=2, b=n$ Apr 6, 2015 at 19:25

Since $2^{2n}=4^n$, you have $\frac{3^n}{2^{2n}}=\left(\frac34\right)^n\rightarrow 0$.

• What exponent rule did you use to simplify $2^{2n}$ to $4^n$? Apr 6, 2015 at 19:24
• $(a^b)^c=a^{bc}$. Apr 6, 2015 at 19:27

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}