Question
Evaluate the following limits: $\lim _{x \rightarrow 2}\left[\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right]$
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Answer
$ \lim _{x \rightarrow 2}\left[\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right]$
$=\lim _{x \rightarrow 2}\left[\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}} \times \frac{\sqrt{x+2}+\sqrt{3 x-2}}{\sqrt{x+2}+\sqrt{3 x-2}}\right]......[$ By rationalization $]$
$=\lim _{x \rightarrow 2} \frac{\left(x^2-4\right)(\sqrt{x+2}+\sqrt{3 x-2})}{(x+2)-(3 x-2)}$
$=\lim _{x \rightarrow 2} \frac{\left(x^2-4\right)(\sqrt{x+2}+\sqrt{3 x-2})}{-2 x+4}$
$=\lim _{x \rightarrow 2} \frac{(x+2)(x-2)(\sqrt{x+2}+\sqrt{3 x-2})}{-2(x-2)}$
$=\lim _{x \rightarrow 2} \frac{(x+2)(\sqrt{x+2}+\sqrt{3 x-2})}{-2}$
$=\frac{(2+2)\left(\because x \rightarrow 2, x \neq 2$
$\therefore x-2 \neq 0\right]}{-2}$
$=\frac{4(2+2)}{-2}$
$=\frac{16}{-2}$
$=-8 $
$=\lim _{x \rightarrow 2}\left[\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}} \times \frac{\sqrt{x+2}+\sqrt{3 x-2}}{\sqrt{x+2}+\sqrt{3 x-2}}\right]......[$ By rationalization $]$
$=\lim _{x \rightarrow 2} \frac{\left(x^2-4\right)(\sqrt{x+2}+\sqrt{3 x-2})}{(x+2)-(3 x-2)}$
$=\lim _{x \rightarrow 2} \frac{\left(x^2-4\right)(\sqrt{x+2}+\sqrt{3 x-2})}{-2 x+4}$
$=\lim _{x \rightarrow 2} \frac{(x+2)(x-2)(\sqrt{x+2}+\sqrt{3 x-2})}{-2(x-2)}$
$=\lim _{x \rightarrow 2} \frac{(x+2)(\sqrt{x+2}+\sqrt{3 x-2})}{-2}$
$=\frac{(2+2)\left(\because x \rightarrow 2, x \neq 2$
$\therefore x-2 \neq 0\right]}{-2}$
$=\frac{4(2+2)}{-2}$
$=\frac{16}{-2}$
$=-8 $
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