Question

Evaluate:$\underset{x\to 0}{\lim }\frac{\sqrt{a+x}-\sqrt{a-x}}{4x}$

• A. $\frac{1}{2\sqrt{a}}$
• B. $\frac{1}{4\sqrt{a}}$
• C. $\frac{1}{3\sqrt{a}}$
• D. $\frac{2}{\sqrt{a}}$

Answer 1

The correct option is B

$\frac{1}{4\sqrt{a}}$

Find the value of $\underset{x\to 0}{\lim }\frac{\sqrt{a+x}-\sqrt{a-x}}{4x}$

Consider the given value as,

$$\begin{gathered} I=\underset{x\to 0}{\lim }\frac{\sqrt{a+x}-\sqrt{a-x}}{4x} \\ \Rightarrow I=\frac{\sqrt{a+0}-\sqrt{a-0}}{4\left(0\right)} \\ \Rightarrow I=\frac{\sqrt{a}-\sqrt{a}}{0} \\ \Rightarrow I=\frac{0}{0}\left[\text{Indeterminent form}\right] \end{gathered}$$

Using L' hospital rule

$\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$

$$\begin{gathered} \Rightarrow I=\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{1}{2\sqrt{a+x}}}-{\displaystyle \frac{1\times -1}{2\sqrt{a-x}}}}{4\times 1} \\ \Rightarrow I=\frac{{\displaystyle \frac{1}{2\sqrt{a+0}}-\frac{-1}{2\sqrt{a-0}}}}{4} \\ \Rightarrow I=\frac{{\displaystyle \frac{1}{2\sqrt{a}}+\frac{1}{2\sqrt{a}}}}{4} \\ \Rightarrow I=\frac{1}{4\sqrt{a}} \\ \Rightarrow I=\frac{1}{4\sqrt{a}} \end{gathered}$$

$\underset{x\to 0}{\lim }\frac{\sqrt{a+x}-\sqrt{a-x}}{4x}=\frac{1}{4\sqrt{a}}$

Hence, the correct answer is option B.