Question

Find the value of $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x+2\sin x}{\sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x-{{\sin }^{2}}x+1}}$ .
(a) 2
(b) 1
(c) 6
(d) $-2$

Answer 1

In this question, we have to evaluate the limit $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x+2\sin x}{\sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x-{{\sin }^{2}}x+1}}$. We will use the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ in the expression $\dfrac{x+2\sin x}{\sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x-{{\sin }^{2}}x+1}}$ to simplify the problem. We use the property of rationalizing a function say $\dfrac{x+a}{x+b}$ by multiplying and dividing the expression by $x-b$. Using this we will then rationalize the function
$\dfrac{x+2\sin x}{\sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x-{{\sin }^{2}}x+1}}$ by multiplying and dividing the expression $\sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x-{{\sin }^{2}}x+1}$ to further simplify the problem. We will also use the identity $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$. Also we will be using the $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$. We will then solve to get the desired answer.

Complete step by step answer:
We are given with the limit $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x+2\sin x}{\sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x-{{\sin }^{2}}x+1}}$.
Consider the expression $\dfrac{x+2\sin x}{\sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x-{{\sin }^{2}}x+1}}$ whose limit is to be calculated when $x$ tends to zero.
Since we know the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, that implies
$1-{{\sin }^{2}}x={{\cos }^{2}}x$.
Thus on substituting the value $1-{{\sin }^{2}}x={{\cos }^{2}}x$ in the given expression $\dfrac{x+2\sin x}{\sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x-{{\sin }^{2}}x+1}}$, we will have
$\dfrac{x+2\sin x}{\sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x-{{\sin }^{2}}x+1}}=\dfrac{x+2\sin x}{\sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x+{{\cos }^{2}}x}}$
Now since we know that in order to rationalize a function say $\dfrac{x+a}{x+b}$ by multiplying and dividing the expression by $x-b$.
we will then rationalize the function
$\dfrac{x+2\sin x}{\sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x+{{\cos }^{2}}x}}$ by multiplying and dividing the expression $\sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x}$ to further simplify the problem.
Using this we have
$\dfrac{x+2\sin x}{\sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x+{{\cos }^{2}}x}}=\dfrac{\left( x+2\sin x \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{\left( \sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x+{{\cos }^{2}}x} \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}$since we also know the identity $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$.
Using this in the denominator of the above expression, we will get

& \dfrac{\left( x+2\sin x \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{\left( \sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x+{{\cos }^{2}}x} \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}=\dfrac{\left( x+2\sin x \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{{{\left( \sqrt{{{x}^{2}}+2\sin x+1} \right)}^{2}}-{{\left( \sqrt{x+{{\cos }^{2}}x} \right)}^{2}}} \\\ & =\dfrac{\left( x+2\sin x \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{\left( {{x}^{2}}+2\sin x+1 \right)-\left( x+{{\cos }^{2}}x \right)} \end{aligned}$$ On simplifying the denominator of the above equation by substituting $$1-{{\sin }^{2}}x={{\cos }^{2}}x$$ , we will have $$\begin{aligned} & \dfrac{\left( x+2\sin x \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{{{x}^{2}}+2\sin x+1-\left( x+1-{{\sin }^{2}}x \right)}=\dfrac{\left( x+2\sin x \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{{{x}^{2}}+2\sin x+1-x-1+{{\sin }^{2}}x} \\\ & =\dfrac{\left( x+2\sin x \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{{{x}^{2}}+2\sin x-x+{{\sin }^{2}}x} \end{aligned}$$ Now on dividing the above expression by $$x$$, we will have $$\begin{aligned} & \dfrac{\left( x+2\sin x \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{{{x}^{2}}+2\sin x-x+{{\sin }^{2}}x}=\dfrac{\dfrac{\left( x+2\sin x \right)}{x}\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{\left( \dfrac{{{x}^{2}}+2\sin x-x+{{\sin }^{2}}x}{x} \right)} \\\ & =\dfrac{\left( \dfrac{x}{x}+2\left( \dfrac{\sin x}{x} \right) \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{\left( x+2\left( \dfrac{\sin x}{x} \right)-1+\sin x\left( \dfrac{\sin x}{x} \right) \right)} \\\ & =\dfrac{\left( 1+2\dfrac{\sin x}{x} \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{x+2\left( \dfrac{\sin x}{x} \right)-1+\sin x\left( \dfrac{\sin x}{x} \right)} \end{aligned}$$ Since we know that $$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$$. Therefore by taking the $$\underset{x\to 0}{\mathop{\lim }}\,$$ in the above expression we will get $$\begin{aligned} & \underset{x\to 0}{\mathop{\lim }}\,\left( \dfrac{\left( 1+2\dfrac{\sin x}{x} \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{x+2\left( \dfrac{\sin x}{x} \right)-1+\sin x\left( \dfrac{\sin x}{x} \right)} \right)=\dfrac{\left( 1+2\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x} \right)\left( \underset{x\to 0}{\mathop{\lim }}\,\sqrt{{{x}^{2}}+2\sin x+1}+\underset{x\to 0}{\mathop{\lim }}\,\sqrt{x+{{\cos }^{2}}x} \right)}{\underset{x\to 0}{\mathop{\lim }}\,x+2\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}-\underset{x\to 0}{\mathop{\lim }}\,1+\underset{x\to 0}{\mathop{\lim }}\,\sin x\left( \dfrac{\sin x}{x} \right)} \\\ & =\dfrac{\left( 1+2 \right)\left( 1+1 \right)}{0+2-1+0} \\\ & =\dfrac{6}{1} \\\ & =6 \end{aligned}$$Since $$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x+2\sin x}{\sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x-{{\sin }^{2}}x+1}}=\underset{x\to 0}{\mathop{\lim }}\,\left( \dfrac{\left( 1+2\dfrac{\sin x}{x} \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{x+2\dfrac{\sin x}{x}-1+\sin x\left( \dfrac{\sin x}{x} \right)} \right)$$ Therefore we have $$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x+2\sin x}{\sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x-{{\sin }^{2}}x+1}}=6$$ **So, the correct answer is “Option C”.** **Note:** In this problem, in order to determine the $$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x+2\sin x}{\sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x-{{\sin }^{2}}x+1}}$$ please take care while rationalizing the expression $$\dfrac{x+2\sin x}{\sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x+{{\cos }^{2}}x}}$$ we have multiplying and dividing the expression $$\sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x}$$ and not by $$\sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x+{{\cos }^{2}}x}$$.