Question

Find the value of $\displaystyle \lim_{x \to {{1}^{-}}}\text{ }\dfrac{\sqrt{\pi }-\sqrt{2{{\sin }^{-1}}x}}{\sqrt{1-x}}$

& A.\dfrac{1}{\sqrt{2\pi }} \\\ & B.\sqrt{\dfrac{\pi }{2}} \\\ & C.\sqrt{\dfrac{2}{\pi }} \\\ & D.\sqrt{\pi } \\\ \end{aligned}$$

Answer 1

To solve this question, first we will check by putting limit values that the given function is $\dfrac{0}{0}\Rightarrow \dfrac{\infty }{\infty }$ form or not then if this is the case then, we will apply L.Hospital Rule which allows us to differentiate numerator and denominator of given function separately. In between while differentiating we will use $\dfrac{d}{dx}{{\sin }^{-1}}x=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$
$\dfrac{d}{dx}\left( f\left( x \right)g\left( x \right) \right)=f\left( x \right)\dfrac{d}{dx}\left( g\left( x \right) \right)+g\left( x \right)\dfrac{d}{dx}\left( f\left( x \right) \right)$

Complete step by step answer:
We are given $\displaystyle \lim_{x \to {{1}^{-}}}\text{ }\dfrac{\sqrt{\pi }-\sqrt{2{{\sin }^{-1}}x}}{\sqrt{1-x}}$ let it be I.
We will first try to obtain answer of the question that if after applying limit value are we getting $\dfrac{0}{0}\Rightarrow \dfrac{\infty }{\infty }\text{ form}$ If so then we will apply L.Hospital rule.
L.Hospital rule is a method which is applicable when the obtained value is of type $\dfrac{0}{0}\Rightarrow \dfrac{\infty }{\infty }$
To apply this rule after we get $\dfrac{0}{0}\Rightarrow \dfrac{\infty }{\infty }$ form we just differentiate both numerator and denominator separately with respect to the given function:
Here, we have $\displaystyle \lim_{x \to {{1}^{-}}}\text{ }\dfrac{\sqrt{\pi }-\sqrt{2{{\sin }^{-1}}x}}{\sqrt{1-x}}$
Applying limit $x \to 1$ we get:
$\displaystyle \lim_{x \to {{1}^{-}}}\text{ }\dfrac{\sqrt{\pi }-\sqrt{2{{\sin }^{-1}}x}}{\sqrt{1-x}}=\dfrac{\sqrt{\pi }-\sqrt{2{{\sin }^{-1}}\left( 1 \right)}}{\sqrt{1-1}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}$
Now, we have $\sin \dfrac{\pi }{2}=1$
Applying ${{\sin }^{-1}}$ both sides we get:

& {{\sin }^{-1}}\left( \sin \dfrac{\pi }{2} \right)={{\sin }^{-1}}1 \\\ & \dfrac{\pi }{2}={{\sin }^{-1}}1 \\\ & {{\sin }^{-1}}1=\dfrac{\pi }{2} \\\ \end{aligned}$$ Using this in equation (i) we get: $$\begin{aligned} & \displaystyle \lim_{x \to {{1}^{-}}}\text{ }\dfrac{\sqrt{\pi }-\sqrt{2{{\sin }^{-1}}x}}{\sqrt{1-x}}=\dfrac{\sqrt{\pi }-\sqrt{2\cdot \dfrac{\pi }{2}}}{\sqrt{1-1}} \\\ & \Rightarrow \dfrac{\sqrt{\pi }-\sqrt{\pi }}{\sqrt{1-1}}=\dfrac{0}{0} \\\ \end{aligned}$$ Hence, we have obtained $\dfrac{0}{0}$ form, so we apply L.Hospital rule. Applying L.Hospital rule in $\displaystyle \lim_{x \to {{1}^{-}}}\text{ }\dfrac{\sqrt{\pi }-\sqrt{2{{\sin }^{-1}}x}}{\sqrt{1-x}}=I$ Hence, differentiating both numerator and denominator separately with respect to x we get: $$I=\displaystyle \lim_{x \to {{1}^{-}}}\text{ }\dfrac{\dfrac{d}{dx}\left( \sqrt{\pi }-\sqrt{2{{\sin }^{-1}}x} \right)}{\dfrac{d}{dx}\left( \sqrt{1-x} \right)}$$ Now, we have $$\dfrac{d}{dx}\left( f\left( x \right)-g\left( x \right) \right)=\dfrac{d}{dx}\left( f\left( x \right) \right)-\dfrac{d}{dx}\left( g\left( x \right) \right)$$ where f and g are functions of x and $\dfrac{d}{dx}{{\sin }^{-1}}x=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ and $\dfrac{d}{dx}\sqrt{1-x}=\dfrac{1\left( -1 \right)}{2\sqrt{1-x}}$ Using this all in above we get: $$\begin{aligned} & I=\displaystyle \lim_{x \to {{1}^{-}}}\text{ }\dfrac{0-\dfrac{1}{2\sqrt{2{{\sin }^{-1}}x}}\dfrac{d}{dx}\left( 2{{\sin }^{-1}}x \right)}{\dfrac{1\left( 1 \right)}{2\sqrt{1-x}}} \\\ & I=\displaystyle \lim_{x \to {{1}^{-}}}\text{ }\dfrac{0-\dfrac{1}{2\sqrt{2{{\sin }^{-1}}x}}2\dfrac{1}{\sqrt{1-{{x}^{2}}}}}{\dfrac{-1}{2\sqrt{1-x}}} \\\ \end{aligned}$$ When solving $\dfrac{d}{dx}\left( 2{{\sin }^{-1}}x \right)$ we have used product rule of differentiation which is given as below. $$\dfrac{d}{dx}\left( f\left( x \right)g\left( x \right) \right)=f\left( x \right)\dfrac{d}{dx}\left( g\left( x \right) \right)+g\left( x \right)\dfrac{d}{dx}\left( f\left( x \right) \right)$$ Where f and g are functions of x. Applying this in $\dfrac{d}{dx}\left( 2{{\sin }^{-1}}x \right)$ we get: $$\begin{aligned} & \dfrac{d}{dx}\left( 2{{\sin }^{-1}}x \right)=2\dfrac{d}{dx}{{\sin }^{-1}}x+{{\sin }^{-1}}x\dfrac{d}{dx}2 \\\ & \dfrac{d}{dx}\left( 2{{\sin }^{-1}}x \right)=2\dfrac{1}{\sqrt{1-{{x}^{2}}}}+0=\dfrac{2}{\sqrt{1-{{x}^{2}}}} \\\ \end{aligned}$$ So, we have obtained $$I=\displaystyle \lim_{x \to {{1}^{-}}}\text{ }\dfrac{\dfrac{1}{\sqrt{2{{\sin }^{-1}}x}\sqrt{1-{{x}^{2}}}}}{\dfrac{+1}{2\sqrt{1-x}}}$$ Taking $2\sqrt{1-x}$ on numerator we get: $$I=\displaystyle \lim_{x \to {{1}^{-}}}\text{ }\dfrac{2\sqrt{1-x}}{\sqrt{2{{\sin }^{-1}}x}\sqrt{1-{{x}^{2}}}}$$ Using formula ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ in $\left( 1-{{x}^{2}} \right)$ we have $$\left( 1-{{x}^{2}} \right)=\left( 1+x \right)\left( 1-x \right)$$ Taking square root $$\sqrt{\left( 1-{{x}^{2}} \right)}=\sqrt{\left( 1-x \right)\left( 1+x \right)}$$ Using this value of I: $$I=\displaystyle \lim_{x \to {{1}^{-}}}\text{ }\dfrac{2\sqrt{1-x}}{\sqrt{2{{\sin }^{-1}}x}\sqrt{\left( 1-x \right)\left( 1+x \right)}}$$ Cancelling $\sqrt{1-x}$ we get: $$I=\displaystyle \lim_{x \to {{1}^{-}}}\text{ }\dfrac{2}{\sqrt{2{{\sin }^{-1}}x}\sqrt{\left( 1+x \right)}}$$ Cancelling $\sqrt{2}$ from $\sqrt{2}$ after writing $2=\sqrt{2}\sqrt{2}$ we get: $$I=\displaystyle \lim_{x \to {{1}^{-}}}\text{ }\dfrac{\sqrt{2}\sqrt{2}}{\sqrt{2}\sqrt{{{\sin }^{-1}}x}\sqrt{\left( 1+x \right)}}=\displaystyle \lim_{x \to {{1}^{-}}}\text{ }\dfrac{\sqrt{2}}{\sqrt{{{\sin }^{-1}}x\left( 1+x \right)}}$$ Now, finally applying $x \to 1$ we get: $$\begin{aligned} & \text{I=}\dfrac{\sqrt{2}}{\sqrt{{{\sin }^{-1}}1\left( 1+1 \right)}} \\\ & I=\dfrac{\sqrt{2}}{\sqrt{\left( \dfrac{\pi }{2} \right)\left( 2 \right)}}=\dfrac{\sqrt{2}}{\sqrt{\pi }} \\\ \end{aligned}$$ We had before as ${{\sin }^{-1}}(1)=\dfrac{\pi }{2}$ $$I=\sqrt{\dfrac{2}{\pi }}$$ **So, the correct answer is “Option C”.** **Note:** The form $\dfrac{0}{0}\And \dfrac{\infty }{\infty }$ forms are called indeterminate forms. While solving this type of questions, first check after putting a limit value that $\dfrac{0}{0}\Rightarrow \dfrac{\infty }{\infty }\text{ form}$ is achieved. If this is the case then only we can apply the L-Hospital rule. Also, a key point to note in this question is that we always try to obtain a numerator or denominator from x (a variable) before applying limit value. Like here we have got $\sqrt{2}$ in numerator and then we applied the limit.