Question

The value of $\underset{x\to \infty }{\lim }$ $\sqrt{a^2{x}^2+ax+1}-\sqrt{a^2{x}^2+1}$ is

• A. $\frac{1}{2}$
• B. $1$
• C. $2$
• D. None of these

Answer 1

Answer: (A)

Given $\underset{x\to \infty }{\lim }$ $\sqrt{a^2{x}^2+ax+1}-\sqrt{a^2{x}^2+1}$
$=\underset{x\to \infty }{\lim }$ $\frac{a^2{x}^2+ax+1-a^2{x}^2-1}{\sqrt{a^2{x}^2+ax+1}+\sqrt{a^2{x}^2+1}}$
$=\underset{x\to \infty }{\lim }$ $\frac{ax}{x\left[\sqrt{a^2+\frac{a}{x}+\frac{1}{x^2}}+\sqrt{a^2+\frac{1}{x^2}}\right]}$
$=\frac{a}{\sqrt{a^2}+\sqrt{a^2}}=\frac{a}{2a}$
$=\frac{1}{2}$