Evaluate the following limits: _ x 2 [ 2+x-6-x x-2 ]

Question

Evaluate the following limits: $\lim _{x \rightarrow 2}\left[\frac{\sqrt{2+x}-\sqrt{6-x}}{\sqrt{x}-\sqrt{2}}\right]$

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Answer

$\lim _{x \rightarrow 2}\left(\frac{\sqrt{2+x}-\sqrt{6-x}}{\sqrt{x}-\sqrt{2}}\right)$
$=\lim _{x \rightarrow 2}\left[\frac{\sqrt{2+x}-\sqrt{6-x}}{\sqrt{x}-\sqrt{2}} \times \frac{\sqrt{2+x}+\sqrt{6-x}}{\sqrt{2+x}+\sqrt{6-x}} \times \frac{\sqrt{x}+\sqrt{2}}{\sqrt{x}+\sqrt{2}}\right]{\left[\begin{array}{l} \text { By taking conjugates of both, the } \\ \text { numerator as well as the denominator. }
\end{array}\right] } \\
= \lim _{x \rightarrow 2}\left[\frac{(2+x)-(6-x)}{(x-2)} \times \frac{\sqrt{x}+\sqrt{2}}{\sqrt{2+x}+\sqrt{6-x}}\right]$
$= \lim _{x \rightarrow 2}\left[\frac{-4+2 x}{x-2} \times \frac{\sqrt{x}+\sqrt{2}}{\sqrt{2+x}+\sqrt{6-x}}\right]$
$= \lim _{x \rightarrow 2}\left[\frac{2(\sqrt{x}+\sqrt{2})}{x-2} \times \frac{\sqrt{x}+\sqrt{2}}{\sqrt{2+x}+\sqrt{6-x}}\right]$
$\left[\because x \rightarrow 2, x \neq 2,\cdots x-2 \neq 0\right]$
$=\frac{\lim _{x \rightarrow 2} 2(\sqrt{x}+\sqrt{2})}{\lim _{x \rightarrow 2}(\sqrt{2+x}+\sqrt{6-x})}$
$=\frac{2(\sqrt{2}+\sqrt{2})}{\sqrt{2+2}+\sqrt{6-2}}$
$=\frac{2(2 \sqrt{2})}{2+2}$
$=\frac{4 \sqrt{2}}{4}$
$=\sqrt{2}$

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