We saw that $\lim_{n\to\infty} n^3 = \infty$ and $\lim_{n\to\infty} 2^n = \infty$, but as $\tfrac{\infty}{\infty}$ is an indeterminate form we cannot conclude anything about $\textstyle\lim_{n\to\infty} \frac{n^3}{2^n}$.

In this figure, the sequences $a_n = n^3$ (green) and $b_n = 2^n$ (blue) are drawn for $n=1$ up to $n = 12$. From $n = 10$ onwards $a_n$ is smaller than $b_n$, so for the fraction we have $\tfrac{a_n}{b_n} \leq 1$. If the limit of $\tfrac{a_n}{b_n}$ exists it therefore has to lie between $0$ and $1$.

Use this animation to examine the behaviour of $\frac{a_n}{b_n} = \frac{n^3}{2^n}$.

The animation suggests that $\lim_{n\to\infty}\frac{n^3}{2^n}=0$. Intuitively, the denominator of this fraction grows much faster than the numerator. Here is again a plot of $a_n$ and $b_n$, now with $n$ from $1$ to $16$. Observe that the blue sequence indeed increases faster than the green sequence.

We add the following to our list of common limits: