Question

Evaluate: $\underset{x\to 0}{\lim }\left(\frac{x}{\sqrt{1+x}-\sqrt{1-x}}\right)$

• A. $0$
• B. $1$
• C. $2$
• D. $-1$
• E. $-2$

Answer 1

The correct option is B

$1$

Explanation for the correct answer:

Given:

$\underset{x\to 0}{\lim }\left(\frac{x}{\sqrt{1+x}-\sqrt{1-x}}\right)$

To find:

The value of $\underset{x\to 0}{\lim }\left(\frac{x}{\sqrt{1+x}-\sqrt{1-x}}\right)$

Now, consider the given Equation as

$I=\underset{x\to 0}{\lim }\left(\frac{x}{\sqrt{1+x}-\sqrt{1-x}}\right)$

Rationalizing the function we get,

$$\begin{gathered} I=\underset{x\to 0}{\lim }\left(\frac{x}{\sqrt{1+x}-\sqrt{1-x}}\right) \\ I=\underset{x\to 0}{\lim }\left(\frac{x(\sqrt{1+x}+\sqrt{1-x})}{(\sqrt{1+x}+\sqrt{1-x})(\sqrt{1+x}-\sqrt{1-x})}\right) \\ I=\underset{x\to 0}{\lim }\left(\frac{x(\sqrt{1+x}+\sqrt{1-x})}{(1+x-1+x)}\right)\left\{\because (a+b)(a-b)=a^2-b^2\right\} \\ I=\underset{x\to 0}{\lim }\left(\frac{x(\sqrt{1+x}+\sqrt{1-x})}{\left(2x\right)}\right) \\ I=\underset{x\to 0}{\lim }\left(\frac{(\sqrt{1+x}+\sqrt{1-x})}{\left(2\right)}\right) \end{gathered}$$

Now, Put directly the value of limit we get,

$$\begin{gathered} I=\underset{x\to 0}{\lim }\left(\frac{(\sqrt{1+x}+\sqrt{1-x})}{\left(2\right)}\right) \\ I=\left(\frac{(\sqrt{1+0}+\sqrt{1-0})}{\left(2\right)}\right) \\ I=1 \end{gathered}$$

Therefore, the value of $\underset{x\to 0}{\lim }\left(\frac{x}{\sqrt{1+x}-\sqrt{1-x}}\right)=1$

Hence, the correct answer is Option B.