Worked exercise — Limits of rational sequences and roots

Problem

Compute the limits: $\text{(a)}\ a_n = \frac{n+1}{n^2+3}\qquad \text{(b)}\ a_n = \frac{2n^2-1}{n^2+n}$ $\text{(c)}\ a_n = \sqrt{n+1}-\sqrt{n}\qquad \text{(d)}\ a_n = \frac{3^n+2^n}{3^n}$

Solution

(a) The denominator ($\text{degree }2$) grows faster than the numerator ($\text{degree }1$): $\frac{n+1}{n^2+3} = \frac{\frac1n + \frac{1}{n^2}}{1+\frac{3}{n^2}} \to \frac{0}{1} = \boxed{0}.$ (b) Same degree above and below: the ratio of the leading coefficients prevails. $\frac{2n^2-1}{n^2+n} = \frac{2-\frac{1}{n^2}}{1+\frac1n} \to \frac{2}{1} = \boxed{2}.$ (c) Form $\infty-\infty$; we rationalise by multiplying by the sum: $\sqrt{n+1}-\sqrt{n} = \frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}} = \frac{1}{\sqrt{n+1}+\sqrt{n}} \to \boxed{0}.$ (d) We split the ratio: $\frac{3^n+2^n}{3^n} = 1 + \left(\frac{2}{3}\right)^n \to 1 + 0 = \boxed{1}$ (because $|2/3|<1$).

Links

Topics: Sequences
Concepts: Algebra of limits · Indeterminate form · Limit of a sequence
Skills: Computing limits
Exercise type: Computing a limit