Question
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\text{x}}$

Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\text{x}}$$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}\big)}{\text{x}}\times\frac{\big(\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}\big)}{\big(\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(1+\text{x}^2\big)-\big(1-\text{x}^2\big)}{\text{x}\big(\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\text{x}^2}{\text{x}\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}^2}\big)}$
$=\frac{2\times0}{\big(\sqrt{1}+\sqrt{1}\big)}$
$=\frac{2}{2}\times0$
$=0$

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