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