Find the value

Question:

Evaluate

$\lim _{x \rightarrow 0}\left(\frac{2 x}{\sqrt{a+x}-\sqrt{a-x}}\right)$

 

Solution:

To evaluate:

$\lim _{x \rightarrow 0} \frac{2 x}{\sqrt{a+x}-\sqrt{a-x}}$

Formula used:

Multiplying numerator and denominator by

$\sqrt{a+x}+\sqrt{a-x}$

$\lim _{x \rightarrow 0} \frac{2 x}{\sqrt{a+x}-\sqrt{a-x}}=\lim _{x \rightarrow 0} \frac{2 x}{\sqrt{a+x}-\sqrt{a-x}}\left(\frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}\right)$

$\lim _{x \rightarrow 0} \frac{2 x}{\sqrt{a+x}-\sqrt{a-x}}=\lim _{x \rightarrow 0} \frac{2 x(\sqrt{a+x}+\sqrt{a-x})}{a+x-a+x}$

$\lim _{x \rightarrow 0} \frac{2 x}{\sqrt{a+x}-\sqrt{a-x}}=\lim _{x \rightarrow 0} \frac{2 x(\sqrt{a+x}+\sqrt{a-x})}{2 x}$

$\lim _{x \rightarrow 0} \frac{2 x}{\sqrt{a+x}-\sqrt{a-x}}=\lim _{x \rightarrow 0} \sqrt{a+x}+\sqrt{a-x}$

$\lim _{x \rightarrow 0} \frac{2 x}{\sqrt{a+x}-\sqrt{a-x}}=2 \sqrt{a}$

Thus, the value of $\lim _{x \rightarrow 0} \frac{2 x}{\sqrt{a+x}-\sqrt{a-x}}$ is $2 \sqrt{a}$.

 

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now