Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+3\text{x}}-\sqrt{1-3\text{x}}}{\text{x}}$

Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+3\text{x}}-\sqrt{1-3\text{x}}}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+3\text{x}}-\sqrt{1-3\text{x}}\big)}{\text{x}}\times\frac{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(1+3\text{x})-(1-3\text{x})}{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{6\text{x}}{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{6}{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}$
$=\frac{6}{\sqrt{1}+\sqrt{1}}$
$=\frac62$
$=3$

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