[Math] How to evaluate this exponential fraction limit

asymptoticscalculusderivativesfunctionslimits

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?

Best Answer

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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