Evaluate the following limit: _ x 0 2- 1+ ^2x

Question

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

✓

Answer

Given that:
$\lim\limits_{\text{x} \rightarrow0}\frac{\sqrt{2}-\sqrt{1+\cos\text{x}}}{\sin^{2}\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\sqrt{2}-\sqrt{1+\cos\text{x}}}{\sin^{2}\text{x}}\times\frac{\sqrt{2}+\sqrt{1+\cos\text{x}}}{\sqrt{2}+\sqrt{1+\cos\text{x}}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{2-(1+\cos\text{x})}{\sin^{2}\text{x}\big[\sqrt{2}+\sqrt{1+\cos\text{x}}\big]}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{1+\cos\text{x}}{\sin^{2}\text{x}\Big[\sqrt{2}+\sqrt{1+\cos\text{x}}\Big]}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin^{2}\frac{\text{x}}{2}}{(2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2})^{2}}\times\frac{1}{\sqrt{2}+\sqrt{1+\cos\text{x}}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin^{2}\frac{\text{x}}{2}}{4\sin^{2}\frac{\text{x}}{2}\cos^{2}\frac{\text{x}}{2}}\times\frac{1}{\sqrt{2}+\sqrt{1+\cos\text{x}}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{2}{4\cos^{2}\frac{\text{x}}{2}}\times\frac{1}{\sqrt{2}+\sqrt{1+\cos\text{x}}}$
Taking limit, we get:
$=\frac{2}{4\cos^{2}0}\times\frac{1}{(\sqrt{2}+\sqrt{2})}$
$=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2\sqrt{2}}=\frac{1}{4\sqrt{2}}$
Hence, the required answer is $\frac{1}{4\sqrt{2}}.$

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