18. Sequences

b5c. Indeterminate Form \(0\cdot\infty\)

Methods:

1. Rewrite the limit as \(\dfrac{0}{0}\) or \(\dfrac{\infty}{\infty}\) and apply a technique for those indeterminate forms.
2. If the factors are differentiable functions of \(n\), it may be useful to change variables from \(n\) to \(t=\dfrac{1}{n}\) and take the limit as \(t\to0^+\).

Compute \(\displaystyle \lim_{n\to\infty}n^2\left[1-\cos\left(\dfrac{1}{n}\right)\right]\).

As \(n\) gets large \(\dfrac{1}{n}\) approaches \(0\), \(\cos\left(\dfrac{1}{n}\right)\) approaches \(1\) and so \(\left[1-\cos\left(\dfrac{1}{n}\right)\right]\) approaches \(0\). So the limit has the indeterminate form \(0\cdot\infty\). We change variables to \(t=\dfrac{1}{n}\) (which changes the indeterminate form to \(\dfrac{0}{0}\)) and apply l'Hopital's Rule: \[\begin{aligned} \lim_{n\to\infty}n^2\left[1-\cos\left(\dfrac{1}{n}\right)\right] &=\lim_{t\to 0^+} \dfrac{1-\cos(t)}{t^2} \\[10pt] \overset{\text{l'H}}{=}\lim_{t\to 0^+} \dfrac{\sin(t)}{2t} &\overset{\text{l'H}}{=}\lim_{t\to 0^+} \dfrac{\cos(t)}{2}=\dfrac{1}{2} \end{aligned}\]