Question
Evaluate:
$\lim\limits_{\text{x} \rightarrow 2}\frac{\text{x}^{2}-4}{\sqrt{3\text{x}-2}-\sqrt{\text{x}+2}}$

Answer

Given that $\lim\limits_{\text{x} \rightarrow 2}\frac{\text{x}^{2}-4}{\sqrt{3\text{x}-2}-\sqrt{\text{x}+2}}$
Rationalizing the denomonator, we get
$=\lim\limits_{\text{x} \rightarrow 2}\frac{(\text{x}+2)(\text{x}+2)\big[\sqrt{3\text{x}-2}+\sqrt{\text{x}+2}\big]}{\big[\sqrt{3\text{x}-2}-\sqrt{\text{x}+2}\big]\big[\sqrt{3\text{x}-2}+\sqrt{\text{x}+2}\big]}$
$=\lim\limits_{\text{x} \rightarrow 2}\frac{(\text{x}+2)(\text{x}+2)\big[\sqrt{3\text{x}-2}+\sqrt{\text{x}+2}\big]}{3\text{x}-2-\text{x}-2}$
$=\lim\limits_{\text{x} \rightarrow 2}\frac{(\text{x}+2)(\text{x}+2)\big[\sqrt{3\text{x}-2}+\sqrt{\text{x}+2}\big]}{2\text{x}-4}$
$=\lim\limits_{\text{x} \rightarrow 2}\frac{(\text{x}+2)(\text{x}+2)\big[\sqrt{3\text{x}-2}+\sqrt{\text{x}+2}\big]}{2(\text{x}-2)}$
$=\lim\limits_{\text{x} \rightarrow 2}\frac{(\text{x}+2)(\text{x}+2)\big[\sqrt{3\text{x}-2}+\sqrt{\text{x}+2}\big]}{2}$
Taking limits, we have
$=\frac{(2+2)\big[\sqrt{6-2}+\sqrt{2+2}\big]}{2}=\frac{4\big[2+2\big]}{2}$
$=\frac{4\times4}{2}=8$
Hence, the required answer is 8.

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