Question

Evaluate$\underset{x\to 0}{\lim }\frac{{\sin }^{2}x}{(\sqrt{2}-\sqrt{1+\cos x})}$:

• A. $\sqrt{2}$
• B. $2$
• C. $4$
• D. $4\sqrt{2}$

Answer 1

The correct option is D

$4\sqrt{2}$

Explanation for correct option

Evaluating the given limit expression

Given, the expression is: $\underset{x\to 0}{\lim }\frac{{\sin }^{2}x}{(\sqrt{2}-\sqrt{1+\cos x})}$

Solving we get,

$$\begin{gathered} =\underset{x\to 0}{\lim }\frac{{\sin }^{2}x}{(\sqrt{2}-\sqrt{1+\cos x})}, \\ =\underset{x\to 0}{\lim }\frac{{\sin }^{2}x}{(\sqrt{2}-\sqrt{1+\cos x)}}\times \frac{(\sqrt{2}+\sqrt{1+\cos x})}{(\sqrt{2}+\sqrt{1+\cos x})} \\ =\underset{x\to 0}{\lim }\frac{\left({\sin }^{2}x\right)}{1}\times \frac{(\sqrt{2}+\sqrt{1+\cos x})}{({\left(\sqrt{2}\right)}^2-{\left(\sqrt{1+\cos x}\right)}^{2)}}[\mathbf{\because }\mathbf{}\mathbf{}\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathbf{\left)}\mathbf{\right(}\mathit{a}\mathbf{-}\mathit{b}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}{\mathit{a}}^{\mathbf{2}}\mathbf{-}{\mathit{b}}^{\mathbf{2}}\mathbf{]} \\ =\underset{x\to 0}{\lim }\frac{\left({\sin }^{2}x\right)\times (\sqrt{2}+\sqrt{1+\cos x})}{(2-(1+\cos x\left)\right)} \\ =\underset{x\to 0}{\lim }\frac{\left({\sin }^{2}x\right)\times (\sqrt{2}+\sqrt{1+\cos x})}{(1-\cos x)} \\ =\underset{x\to 0}{\lim }\frac{\left({\sin }^{2}x\right)}{1}\times \frac{x^2}{x^2}\times \frac{(\sqrt{2}+\sqrt{1+\cos x})}{(1-\cos x)} \\ =\underset{x\to 0}{\lim }\frac{\left({\sin }^{2}x\right)}{x^2}\times \frac{x^2}{1}\times \frac{1}{2{\sin }^2\left({\displaystyle \frac{x}{2}}\right)}\times (\sqrt{2}+\sqrt{1+\cos x})[\mathbf{\because }\mathbf{}\mathbf{(}\mathbf{1}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{2}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{x}}{\mathbf{2}}\mathbf{\right)}\mathbf{\left)}\mathbf{\right]} \\ =\underset{x\to 0}{\lim }\frac{\left({\sin }^{2}x\right)}{x^2}\times \frac{x^2}{1}\times \frac{1}{2{\sin }^2\left({\displaystyle \frac{x}{2}}\right)}\times (\sqrt{2}+\sqrt{1+\cos x}) \\ =\underset{x\to 0}{\lim }\frac{\left({\sin }^{2}x\right)}{x^2}\times \left(\frac{x^2}{4}\right)\times \frac{4}{2{\sin }^2\left({\displaystyle \frac{x}{2}}\right)}\times (\sqrt{2}+\sqrt{1+\cos x}) \\ =\underset{x\to 0}{\lim }\frac{\left({\sin }^{2}x\right)}{x^2}\times \underset{x\to 0}{\lim }\left(\frac{1}{{\displaystyle \frac{{\sin }^2\left(\frac{x}{2}\right)}{\left(\frac{x^2}{4}\right)}}}\right)\times \frac{4}{2}\times \underset{x\to 0}{\lim }(\sqrt{2}+\sqrt{1+\cos x}) \\ L=\underset{x\to 0}{\lim }{\left(\frac{\sin x}{x}\right)}^2\times \underset{x\to 0}{\lim }\left(\frac{1}{{\displaystyle {\left(\frac{\sin \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)}\right)}^2}}\right)\times 2\times \underset{x\to 0}{\lim }(\sqrt{2}+\sqrt{1+\cos x}) \\ L=1\times 1\times 2\times 2\sqrt{2}[\mathbf{\because }\mathbf{}\underset{\mathbf{t}\mathbf{\to }\mathbf{0}}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{t}}{\mathbf{t}}\mathbf{}\mathbf{=}\mathbf{1}] \\ L=4\sqrt{2} \end{gathered}$$

Hence, the correct answer is option (D).