Find the value



$\lim _{x \rightarrow 2}\left(\frac{x^{2}-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)$



To evaluate:

$\lim _{x \rightarrow 2} \frac{x^{2}-4}{\sqrt{x+2}-\sqrt{3 x-2}}$

Formula used: L'Hospital's rule

Let $f(x)$ and $g(x)$ be two functions which are differentiable on an open interval I except at a point a where

$\lim _{x \rightarrow a} \mathrm{f}(\mathrm{x})=\lim _{x \rightarrow a} \mathrm{~g}(\mathrm{x})=0$ or $\pm \infty$


$\lim _{x \rightarrow a} \frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}=\lim _{x \rightarrow a} \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{g}^{\prime}(\mathrm{x})}$

As $x \rightarrow 0$, we have

$\lim _{x \rightarrow 2} \frac{x^{2}-4}{\sqrt{x+2}-\sqrt{3 x-2}}=\frac{0}{0}$

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

$\lim _{x \rightarrow 2} \frac{x^{2}-4}{\sqrt{x+2}-\sqrt{3 x-2}}=\lim _{x \rightarrow 2} \frac{\frac{d}{d x}\left(x^{2}-4\right)}{d x}(\sqrt{x+2}-\sqrt{3 x-2})$

$\lim _{x \rightarrow 2} \frac{x^{2}-4}{\sqrt{x+2}-\sqrt{3 x-2}}=\lim _{x \rightarrow 2} \frac{1}{2 \sqrt{x+2}-\frac{2 x}{2 \sqrt{3 x-2}}}$

$\lim _{x \rightarrow 2} \frac{x^{2}-4}{\sqrt{x+2}-\sqrt{3 x-2}}=\frac{4}{\frac{1}{2 \sqrt{2+2}}-\frac{3}{2 \sqrt{6-2}}}$

$\lim _{x \rightarrow 2} \frac{x^{2}-4}{\sqrt{x+2}-\sqrt{3 x-2}}=\frac{8}{\frac{1}{2}-\frac{3}{2}}$

$\lim _{x \rightarrow 2} \frac{x^{2}-4}{\sqrt{x+2}-\sqrt{3 x-2}}=-8$

Thus, the value of $\lim _{x \rightarrow 2} \frac{x^{2}-4}{\sqrt{x+2}-\sqrt{3 x-2}}$ is $-8$.


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