Question

Evaluate the following limit: $\displaystyle \lim_{x \to 0}\text{ }\dfrac{{{\sin }^{2}}x}{\sqrt{2}-\sqrt{1+\cos x}}$

& \text{A}.\text{ 2}\sqrt{2} \\\ & \text{B}.\text{ 4}\sqrt{2} \\\ & \text{C}.\text{ }\sqrt{2} \\\ & \text{D}.\text{ 4} \\\ \end{aligned}$$

Answer 1

This question is consisting the elementary concept of LIMIT. Here, we see what happens, when we apply the limit $x \to 0$ then the given expression behaves like $\left( \dfrac{0}{0} \right)$ indeterminant form. Now further, we know that, in such types of questions, we should try to remove the term which causes such an indeterminant form. Now, after removing such constraints using rationalization, we get a normal trigonometric expression. These expressions can be reduced to a simpler form using some trigonometric identities. Now, we are able to put our limit to this final expression and get the appropriate answer.

Complete step-by-step solution:
Now, coming to the question, i.e.

& \displaystyle \lim_{x \to 0}\text{ }\dfrac{{{\sin }^{2}}x}{\sqrt{2}-\sqrt{1+\cos x}} \\\ & \Rightarrow \dfrac{{{\sin }^{2}}0}{\sqrt{2}-\sqrt{1+\cos 0}}=\dfrac{0}{\sqrt{2}-\sqrt{2}}=\left( \dfrac{0}{0} \right) \\\ \end{aligned}$$ Look above, we are getting $\left( \dfrac{0}{0} \right)$ indeterminant form like this, we may also get some other indeterminant forms in some other questions. There are basically 7 indeterminant forms $$\left( \dfrac{0}{0},\dfrac{\infty }{\infty },\infty -\infty ,0\times \infty ,{{1}^{\infty }},{{0}^{0}},{{\infty }^{0}} \right)$$ Now, for rationalizing, see in denominator, we have $\left( \sqrt{2}-\sqrt{1+\cos x} \right)$ hence multiply $\left( \sqrt{2}+\sqrt{1+\cos x} \right)$ in numerator and denominator.

\displaystyle \lim_{x \to 0}\text{ }\dfrac{\left( {{\sin }^{2}}x \right)\left( \sqrt{2}+\sqrt{1+\cos x} \right)}{\left( \sqrt{2}-\sqrt{1+\cos x} \right)\left( \sqrt{2}+\sqrt{1+\cos x} \right)}\text{ (rationalise the equation)} \\
\Rightarrow \displaystyle \lim_{x \to 0}\text{ }\dfrac{\left( {{\sin }^{2}}x \right)\left( \sqrt{2}+\sqrt{1+\cos x} \right)}{2-\left( 1+\cos x \right)}\text{ (we know that }\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}) \\
\Rightarrow \displaystyle \lim_{x \to 0}\text{ }\dfrac{\left( {{\sin }^{2}}x \right)\left( \sqrt{2}+\sqrt{1+\cos x} \right)}{\left( 1-\cos x \right)} \\
\Rightarrow \displaystyle \lim_{x \to 0}\text{ }\dfrac{\left( {{\sin }^{2}}x \right)\left( \sqrt{2}+\sqrt{1+\cos x} \right)}{\left( 1-\cos x \right)}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (a)} \\

Now, we know that: $$\begin{aligned} & i)\cos 2x=2{{\cos }^{2}}x-1 \\\ & ii)\cos 2x=1-2{{\sin }^{2}}x \\\ \end{aligned}$$ We use (ii) identity in our question, $$\cos 2x=1-2{{\sin }^{2}}x$$ Replace $x \to \dfrac{x}{2}$ hence, $$\begin{aligned} &\Rightarrow\cos 2\left( \dfrac{x}{2} \right)=1-2{{\sin }^{2}}\left( \dfrac{x}{2} \right) \\\ &\Rightarrow \cos x=1-2{{\sin }^{2}}\dfrac{x}{2} \\\ &\Rightarrow \left( 1-\cos x \right)=2{{\sin }^{2}}\dfrac{x}{2}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (b)} \\\ \end{aligned}$$ Also, we know $$\sin 2x=2\sin x\cos x$$ Replace $x \to \dfrac{x}{2}$ hence, $$\begin{aligned} & \sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \\\ &\Rightarrow {{\sin }^{2}}x=4{{\sin }^{2}}\dfrac{x}{2}{{\cos }^{2}}\dfrac{x}{2}\text{ (take both side square) }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (c)} \\\ \end{aligned}$$ Now, put equation (b) and equation (c) in equation (a):

\displaystyle \lim_{x \to 0}\text{ }\dfrac{\left( 4{{\sin }^{2}}\dfrac{x}{2}{{\cos }^{2}}\dfrac{x}{2} \right)\left( \sqrt{2}+\sqrt{1+\cos x} \right)}{\left( 2{{\sin }^{2}}\dfrac{x}{2} \right)} \\
\Rightarrow\displaystyle \lim_{x \to 0}\text{ 2}{{\cos }^{2}}\dfrac{x}{2}\left( \sqrt{2}+\sqrt{1+\cos x} \right) \\
\Rightarrow \displaystyle \lim_{x \to 0}\text{ }2\times \left( \sqrt{2}+\sqrt{1+1} \right) \\
\Rightarrow \displaystyle \lim_{x \to 0}\text{ }2\times 2\sqrt{2}\text{ }\left( \because \cos 0=1 \right) \\
\Rightarrow \displaystyle \lim_{x \to 0}\text{ }4\sqrt{2} \\ $$
Therefore, the correct option is B $4\sqrt{2}$.

Note: In the above question, we should be aware that, whenever we put the value of x in a trigonometric function, it will be in radian. In this question, we should also know the basics or elementary idea of trigonometric identities otherwise, questions will definitely get stuck in the middle portion. The questions are framed in such a way that could be able to check the multiple concepts. Sometimes we can also use the concept of series expansion (binomial, exponential, logarithmic, sin x, cos x, tan x) as far as the limit question is concerned.