Question

$\underset{x\to \infty }{\lim }\frac{\sqrt{x^2-1}}{2x+1}$ is equal to

(a) 1
(b) 0
(c) −1
(d) 1/2

Answer 1

(d) 1/2

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{\sqrt{x^2-1}}{2x+1} \\ =\underset{x\to \infty }{\lim }\frac{\sqrt{1-{\displaystyle \frac{1}{x^2}}}}{2+{\displaystyle \frac{1}{x}}} \\ \mathrm{Divid}\text{ing the}\mathrm{numerator}\mathrm{and}\text{the}\mathrm{denominator}\mathrm{by}x,\mathrm{we}\mathrm{get}: \\ =\frac{\sqrt{1-0}}{2+0} \\ =\frac{1}{2} \end{gathered}$$

Hence, the correct answer is d.