# Finding the limit of a function $n^3 / 3^n$

$$\lim_{n→\infty} \frac{n^3}{3^n} =0$$

The answer is 0 but how would i go about proving that?

• Use the limit comparison test, surely $2^n > n^3$ for a large enough value of $n$. Apr 9, 2014 at 22:04

Let $a_n = \frac{n^3}{3^n}$. Then

$$\left| \frac{ a_{n+1}}{a_n} \right| = \frac{(n+1)^3}{3^{n+1}} \frac{3^n}{n^3} = \frac{1}{3} \left( \frac{n+1}{n} \right)^3 \to \frac{1}{3} < 1$$

Therefore $a_n \to 0$

Apply l´Hopital's rule three times to the function $f(x)=x^3/3^x$:

$$\lim_{x\rightarrow\infty}\left(\frac{x^3}{3^x}\right)=\lim_{x\rightarrow\infty}\left(\frac{6}{(\ln 3)^3 3^x}\right)=0$$

Note that $n^3 \leq 2^n$ for sufficiently large $n$. Thus, we can bound $a_n = n^3/3^n$ by

$$0 \leq \frac{n^3}{3^n} \leq \frac{2^n}{3^n}.$$

Furthermore, $\lim_{n \to \infty} (2/3)^n = 0$ and so we have

$$0 \leq \lim_{n \to \infty} \frac{n^3}{3^n} \leq 0.$$

By the squeeze theorem, the limit is $0$.

$$\lim_{n\to \infty} \ln(\frac{n^3}{3^n}) = \lim_{n\to \infty} \ln(n^3) - \ln({3^n}) = \lim_{n\to \infty} 3\ln(n) - n\ln({3})=-\infty$$
This is clear because $$\ln(n), so for large enough $$n$$ we have the second term dominating.
Now $$\ln(\cdot)$$ is continuous so $$0=e^{-\infty}=e^{\lim_{n\to \infty} \ln(\frac{n^3}{3^n})} = e^{\ln \lim_{n\to \infty} \frac{n^3}{3^n}} = \lim_{n\to \infty} \frac{n^3}{3^n}$$
• If we can say $n$ dominates over $\ln n$, why not just say that $3^n$ dominates over $n^3$?
• Why can you assume that $L$ exists? If it doesn't, then your conclusion is incorrect. Apr 9, 2014 at 22:21
• Well..... perhaps some things are just more obvious than others. Both are obvious to me, I'm just giving an alternate way of thinking about it. It might be the case that many people just simply confuse $3^n$ and $n^3$; but I doubt anyone would argue with the comparison between $n$ and $\ln(n)$. Just my thought at least Apr 9, 2014 at 22:21