### Limits part 1: Infinite sequences: Standard limits

### Polynomials versus exponential functions

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 the next 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 the animation below 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. Below 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:

For all \(k> 0\) and \(g>1\) we have \[ \lim_{n\to\infty} \frac{n^k}{g^n} =0 \text.\]