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

Hint: 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
\[\begin{align}
  & \dfrac{\left( x+2\sin x \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{\left( \sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x+{{\cos }^{2}}x} \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}=\dfrac{\left( x+2\sin x \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{{{\left( \sqrt{{{x}^{2}}+2\sin x+1} \right)}^{2}}-{{\left( \sqrt{x+{{\cos }^{2}}x} \right)}^{2}}} \\
 & =\dfrac{\left( x+2\sin x \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{\left( {{x}^{2}}+2\sin x+1 \right)-\left( x+{{\cos }^{2}}x \right)}
\end{align}\]
On simplifying the denominator of the above equation by substituting \[1-{{\sin }^{2}}x={{\cos }^{2}}x\] , we will have
\[\begin{align}
  & \dfrac{\left( x+2\sin x \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{{{x}^{2}}+2\sin x+1-\left( x+1-{{\sin }^{2}}x \right)}=\dfrac{\left( x+2\sin x \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{{{x}^{2}}+2\sin x+1-x-1+{{\sin }^{2}}x} \\
 & =\dfrac{\left( x+2\sin x \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{{{x}^{2}}+2\sin x-x+{{\sin }^{2}}x}
\end{align}\]
Now on dividing the above expression by \[x\], we will have
\[\begin{align}
  & \dfrac{\left( x+2\sin x \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{{{x}^{2}}+2\sin x-x+{{\sin }^{2}}x}=\dfrac{\dfrac{\left( x+2\sin x \right)}{x}\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{\left( \dfrac{{{x}^{2}}+2\sin x-x+{{\sin }^{2}}x}{x} \right)} \\
 & =\dfrac{\left( \dfrac{x}{x}+2\left( \dfrac{\sin x}{x} \right) \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{\left( x+2\left( \dfrac{\sin x}{x} \right)-1+\sin x\left( \dfrac{\sin x}{x} \right) \right)} \\
 & =\dfrac{\left( 1+2\dfrac{\sin x}{x} \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{x+2\left( \dfrac{\sin x}{x} \right)-1+\sin x\left( \dfrac{\sin x}{x} \right)}
\end{align}\]
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
\[\begin{align}
  & \underset{x\to 0}{\mathop{\lim }}\,\left( \dfrac{\left( 1+2\dfrac{\sin x}{x} \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{x+{{\cos }^{2}}x} \right)}{x+2\left( \dfrac{\sin x}{x} \right)-1+\sin x\left( \dfrac{\sin x}{x} \right)} \right)=\dfrac{\left( 1+2\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x} \right)\left( \underset{x\to 0}{\mathop{\lim }}\,\sqrt{{{x}^{2}}+2\sin x+1}+\underset{x\to 0}{\mathop{\lim }}\,\sqrt{x+{{\cos }^{2}}x} \right)}{\underset{x\to 0}{\mathop{\lim }}\,x+2\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}-\underset{x\to 0}{\mathop{\lim }}\,1+\underset{x\to 0}{\mathop{\lim }}\,\sin x\left( \dfrac{\sin x}{x} \right)} \\
 & =\dfrac{\left( 1+2 \right)\left( 1+1 \right)}{0+2-1+0} \\
 & =\dfrac{6}{1} \\
 & =6
\end{align}\]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 \[\sqrt{{{x}^{2}}+2\sin x+1}-\sqrt{x+{{\cos }^{2}}x}\].