18. Sequences

b5b. Indeterminate Forms \(\dfrac{0}{0}\) and \(\dfrac{\infty}{\infty}\)

Methods:

1. If the denominator is a sum, divide the numerator and denominator by the largest term in the denominator.
2. If the numerator and denominator are differentiable functions of \(n\), apply l'Hopital's Rule.
3. If the numerator and denominator 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 \(\lim\limits_{n\to\infty}\dfrac{\ln n}{n}\).

This has the indeterminate form \(\dfrac{\infty}{\infty}\). Notice that \(f(x)=\dfrac{\ln x}{x}\) is a differentiable function and our sequence is \(a_n=\dfrac{\ln n}{n}=f(n)\). So we use l'Hopital's Rule: \[ \lim_{n\to\infty}\dfrac{\ln n}{n} =\lim_{x\to\infty}\dfrac{\ln x}{x} \,\overset{\text{l'H}}{=}\, \lim_{x\to\infty}\dfrac{\dfrac{1}{x}}{1}=\dfrac{0}{1}=0 \]

In practice, we do not switch variables and simply regard \(n\) as a continuous variable: \[ \lim_{n\to\infty}\dfrac{\ln n}{n} \,\overset{\text{l'H}}{=}\, \lim_{n\to\infty}\dfrac{\dfrac{1}{n}}{1}=\dfrac{0}{1}=0 \]

Compute \(\lim\limits_{n\to\infty}\dfrac{3n^2}{2n^2+(-1)^n n}\).

This has the indeterminate form \(\dfrac{\infty}{\infty}\). Since \(\dfrac{3n^2}{2n^2+(-1)^n n}\) does not look like a continuous function of \(n\), we cannot use l'Hopital's Rule. So we divide the numerator and denominator by \(n^2\): \[\begin{aligned} \lim_{n\to\infty}\dfrac{3n^2}{2n^2+(-1)^n n} &=\lim_{n\to\infty}\dfrac{3n^2}{2n^2+(-1)^n n}\cdot \dfrac{\dfrac{1}{n^2}}{\dfrac{1}{n^2}} \\[20pt] &=\lim_{n\to\infty}\dfrac{3}{2+\dfrac{(-1)^n}{n}} =\dfrac{3}{2+0}=\dfrac{3}{2} \end{aligned}\]

In fact, we can write \(\dfrac{3n^2}{2n^2+(-1)^n n}\) as a continuous function, if we notice that \((-1)^n=\cos(n\pi)\). However, it is much harder to differentiate \(2n^2+\cos(n\pi)n\) than to divide by \(n^2\).