Prove the following

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

Now to rationalize the denominator by multiplying the given equation by its rationalizing factor we get

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

Taking $(x-2)(x+2)$ as common we get

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

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

Taking $(x-2)(x+2)$ as common we get

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

Taking common and simplifying we get

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

Now by applying the limit we get

$\Rightarrow \lim _{x \rightarrow 2} \frac{(x+2)(\sqrt{3 x-2}+\sqrt{x+2})}{2}=8$

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