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

The correct option is A

$\frac{1}{2}$

Explanation of the correct option.

Compute the required value.

Given : $\underset{x\to \infty }{\lim }\sqrt{a^2{x}^2+ax+1}-\sqrt{a^2{x}^2+1}$

Multiply and divide by $\sqrt{a^2{x}^2+ax+1}+\sqrt{a^2{x}^2+1}$.

$$\begin{gathered} \Rightarrow \underset{x\to \infty }{\lim }\frac{\sqrt{a^2{x}^2+ax+1}-\sqrt{a^2{x}^2+1}}{\sqrt{a^2{x}^2+ax+1}+\sqrt{a^2{x}^2+1}}\times \sqrt{a^2{x}^2+ax+1}+\sqrt{a^2{x}^2+1} \\ \Rightarrow \underset{x\to \infty }{\lim }\frac{\left(a^2{x}^2+ax+1\right)-\left(a^2{x}^2+1\right)}{\sqrt{a^2{x}^2+ax+1}+\sqrt{a^2{x}^2+1}} \\ \Rightarrow \underset{x\to \infty }{\lim }\frac{ax}{ax\sqrt{1+{\displaystyle \frac{ax}{a^2{x}^2}}+{\displaystyle \frac{1}{a^2{x}^2}}}+ax\sqrt{1+\frac{1}{a^2{x}^2}}} \\ \Rightarrow \underset{x\to \infty }{\lim }\frac{ax}{ax\left[\sqrt{1+{\displaystyle \frac{ax}{a^2{x}^2}}+{\displaystyle \frac{1}{a^2{x}^2}}}+\sqrt{1+\frac{1}{a^2{x}^2}}\right]} \\ \Rightarrow \underset{x\to \infty }{\lim }\frac{1}{\left[\sqrt{1+{\displaystyle \frac{a}{a^{2}x}}+{\displaystyle \frac{1}{a^2{x}^2}}}+\sqrt{1+\frac{1}{a^2{x}^2}}\right]} \\ \Rightarrow \underset{x\to \infty }{\lim }\frac{1}{\sqrt{1}+\sqrt{1}} \\ \Rightarrow \frac{1}{2} \end{gathered}$$

Therefore, the value of $\underset{x\to \infty }{\lim }\sqrt{a^2{x}^2+ax+1}-\sqrt{a^2{x}^2+1}$ is $\frac{1}{2}$.

Hence,option $\left(A\right)$ is the correct option.