Question

\[\displaystyle \lim_{x \to 0}\dfrac{{{\sin }^{-1}}x-{{\tan }^{-1}}x}{{{x}^{2}}}\] is equal to:
A. \[{}^{1}/{}_{2}\]
B. \[{}^{-1}/{}_{2}\]
C. \[0\]
D. \[\infty \]

Answer 1

Hint: Solve the given limit to get the form of the limit i.e. whether the given expression is of determinate or indeterminate form. Use L’Hospital Rule to solve the given expression, which states that if the value of \[\displaystyle \lim_{x \to c}\dfrac{f(x)}{g(x)}\] is of the form \[\dfrac{0}{0}or\dfrac{\infty }{\infty }\], then we can differentiate the numerator and denominator individually and hence, can put limits to it. Use the chain rule of differentiation, wherever required, it is given as \[f\left[ g\left( x \right) \right]'=f'\left[ g\left( x \right) \right]\left[ g'\left( x \right) \right]\].

Complete step-by-step answer:
Let the value of the given expression, i.e. \[\displaystyle \lim_{x \to 0}\dfrac{{{\sin }^{-1}}x-{{\tan }^{-1}}}{{{x}^{2}}}\], be M. We get the equation as,
\[M=\displaystyle \lim_{x \to 0}\dfrac{{{\sin }^{-1}}x-{{\tan }^{-1}}}{{{x}^{2}}}.....(1)\]
Let us put limit \[x\to 0\] to the expression of equation and hence we get,
\[M=\displaystyle \lim_{x \to 0}\dfrac{{{\sin }^{-1}}x-{{\tan }^{-1}}}{{{x}^{2}}}\]
\[M=\dfrac{{{\sin }^{-1}}0-{{\tan }^{-1}}0}{{{0}^{2}}}=\dfrac{0-0}{0}=\dfrac{0}{0}\]
where the values of \[{{\sin }^{-1}}0\] and \[{{\tan }^{-1}}0\] are 0.
So we get to know that the given expression is in an indeterminate form of limit. It means that we need to simplify it before putting the limits, so let us apply L’Hospital Rule to the expression, which is defined as if the value of \[\displaystyle \lim_{x \to c}\dfrac{f(x)}{g(x)}\] is of the form \[\dfrac{0}{0}or\dfrac{\infty }{\infty }\], then we can differentiate the numerator and denominator individually and hence, can put limits to it.
Hence we can get the value of M after differentiating the numerator and denominator between as per the L’Hospital rule. So we get,
\[\begin{align}
  & M=\displaystyle \lim_{x \to 0}\dfrac{\dfrac{d}{dx}\left( {{\sin }^{-1}}x-{{\tan }^{-1}}x \right)}{\dfrac{d}{dx}{{x}^{2}}} \\
 & M=\displaystyle \lim_{x \to 0}\dfrac{\dfrac{d}{dx}{{\sin }^{-1}}x-\dfrac{d}{dx}{{\tan }^{-1}}x}{\dfrac{d}{dx}{{x}^{2}}} \\
\end{align}\]
Now we know the relations, given as,
\[\begin{align}
  & \dfrac{d}{dx}{{\sin }^{-1}}x=\dfrac{1}{\sqrt{1-{{x}^{2}}}}.....(2) \\
 & \dfrac{d}{dx}ta{{n}^{-1}}x=\dfrac{1}{1+{{x}^{2}}}.......(3) \\
 & \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}...............(4) \\
\end{align}\]
Hence,
\[\begin{align}
  & M=\dfrac{\dfrac{1}{\sqrt{1-{{x}^{2}}}}-\dfrac{1}{1+{{x}^{2}}}}{2x} \\
 & M=\displaystyle \lim_{x \to 0}\dfrac{{{\left( 1-{{x}^{2}} \right)}^{{}^{-1}/{}_{2}}}-{{\left( 1+{{x}^{2}} \right)}^{-1}}}{2x}......(5) \\
\end{align}\]
Now put \[x\to 0\] to the above expression, so we get,
\[\begin{align}
  & M=\dfrac{{{\left( 1-0 \right)}^{{}^{-1}/{}_{2}}}-{{\left( 1+0 \right)}^{-1}}}{2(0)} \\
 & M=\dfrac{1-1}{0}=\dfrac{0}{0} \\
\end{align}\]
Now we can observe that the expression in the equation (5) is in indeterminate form again. So we need to apply the L'Hospital rule again to the expression. It means we need to differentiate the numerator and denominator again. So we get,
\[\begin{align}
  & M=\displaystyle \lim_{x \to 0}\dfrac{\dfrac{d}{dx}\left[ {{\left( 1-{{x}^{2}} \right)}^{{}^{-1}/{}_{2}}}-{{\left( 1+{{x}^{2}} \right)}^{-1}} \right]}{\dfrac{d}{dx}\left( 2x \right)} \\
 & M=\displaystyle \lim_{x \to 0}\dfrac{\dfrac{d}{dx}{{\left( 1-{{x}^{2}} \right)}^{{}^{-1}/{}_{2}}}-\dfrac{d}{dx}{{\left( 1+{{x}^{2}} \right)}^{-1}}}{2\dfrac{d}{dx}x}......(6) \\
\end{align}\]
Now differentiate the terms \[{{\left( 1-{{x}^{2}} \right)}^{{}^{-1}/{}_{2}}}\] and \[{{\left( 1+{{x}^{2}} \right)}^{-1}}\] with the help of chain rule, which is given as,
\[f\left[ g\left( x \right) \right]'=f'\left[ g\left( x \right) \right]\left[ g'\left( x \right) \right]\].
If we have function of type \[f\left[ g\left( x \right) \right]\], we can get the differentiation of it as, \[f\left[ g\left( x \right) \right]'=f'\left[ g\left( x \right) \right]\left[ g'\left( x \right) \right]........(7)\]
So we can get the value of M from the expression in equation (6) as,
\[\begin{align}
  & M=\displaystyle \lim_{x \to 0}\dfrac{\dfrac{-1}{2}{{\left( 1-{{x}^{2}} \right)}^{{}^{-3}/{}_{2}}}\dfrac{d}{dx}\left( -{{x}^{2}} \right)-(-1){{\left( 1+{{x}^{2}} \right)}^{-2}}\dfrac{d}{dx}{{x}^{2}}}{2} \\
 & M=\displaystyle \lim_{x \to 0}\dfrac{\dfrac{-1}{2}{{\left( 1-{{x}^{2}} \right)}^{{}^{-3}/{}_{2}}}\left( -2x \right)+{{\left( 1+{{x}^{2}} \right)}^{-2}}\left( 2x \right)}{2} \\
 & M=\displaystyle \lim_{x \to 0}\dfrac{x{{\left( 1-{{x}^{2}} \right)}^{{}^{-3}/{}_{2}}}+2x{{\left( 1+{{x}^{2}} \right)}^{-2}}}{2} \\
\end{align}\]
Now put the limit \[x\to 0\] to the above expression, so we get the value of M as,
\[M=\dfrac{0{{\left( 1-0 \right)}^{{}^{-3}/{}_{2}}}+0{{\left( 1+0 \right)}^{-2}}}{2}=\dfrac{0}{2}=0\].
Hence the value of \[\displaystyle \lim_{x \to 0}\dfrac{{{\sin }^{-1}}x-{{\tan }^{-1}}x}{{{x}^{2}}}\] is 0.

So, the correct answer is “Option C”.

Note: Another approach for the question would be that we can put expansion of \[{{\sin }^{-1}}x\] and \[{{\tan }^{-1}}x\] to the given expression.
Expansions are given as,
\[\begin{align}
  & {{\sin }^{-1}}x=x+\dfrac{{{x}^{3}}}{2.3}+\dfrac{1.3{{x}^{5}}}{2.4.5}+........ \\
 & {{\tan }^{-1}}x=x-\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}-\dfrac{{{x}^{7}}}{7}........ \\
\end{align}\]
We can put values of \[{{\sin }^{-1}}x\] and \[{{\tan }^{-1}}x\]to the given expression as well.
We cannot use L’Hospital property with other indeterminate form like \[{{1}^{\infty }},{{0}^{\infty }},{{0}^{0}}\] etc. we can use it only if the limit is of the form \[\dfrac{0}{0}or\dfrac{\infty }{\infty }\].