18. Sequences

b5d. Indeterminate Form \(\infty-\infty\)

The conjugate of a term \(a_n-b_n\) is \(a_n+b_n\).

Methods:

1. Multiply and divide each term by its conjugate. This will convert the indeterminate form to \(\dfrac{\infty}{\infty}\). Then apply a technique for that indeterminate form.
2. Put the terms over a common denominator.

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

We have \[ \lim_{n\to\infty}\dfrac{n^2}{n-1}=\infty \qquad \text{and} \qquad \lim_{n\to\infty}\dfrac{n^2}{n+1}=\infty \] So the limit has the indeterminate form \(\infty-\infty\). We put the terms over a common denominator: \[\begin{aligned} \lim_{n\to\infty} \left(\dfrac{n^2}{n-1}-\dfrac{n^2}{n+1}\right) &=\lim_{n\to\infty} \dfrac{n^2(n+1)-n^2(n-1)}{(n-1)(n+1)} \\ &=\lim_{n\to\infty} \dfrac{2n^2}{n^2-1}=2 \end{aligned}\]

Compute \(\lim\limits_{n\to\infty} \left(\sqrt{n+\sqrt{n}}-\sqrt{n}\right)\).

We multiply and divide by the conjugate, \(\sqrt{n+\sqrt{n}}+\sqrt{n}\): \[\begin{aligned} \lim_{n\to\infty} &\left(\sqrt{n+\sqrt{n}}-\sqrt{n}\right) \\ &=\lim_{n\to\infty} \dfrac{\left(\sqrt{n+\sqrt{n}}-\sqrt{n}\right)\left(\sqrt{n+\sqrt{n}}+\sqrt{n}\right)} {\left(\sqrt{n+\sqrt{n}}+\sqrt{n}\right)} \\ &=\lim_{n\to\infty} \dfrac{n+\sqrt{n}-n} {\sqrt{n+\sqrt{n}}+\sqrt{n}} \\ &=\lim_{n\to\infty} \dfrac{\sqrt{n}} {\sqrt{n+\sqrt{n}}+\sqrt{n}} \\ &=\lim_{n\to\infty} \dfrac{1} {\sqrt{1+\dfrac{1}{\sqrt{n}}}+1}=\dfrac{1}{2} \end{aligned}\] In the next to last step, we divided the numerator and denominator by \(\sqrt{n}\).