Find the value of \[\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \dfrac{\pi }{2}-{{\tan }^{-1}}x \right)}^{{1}/{x}\;}}\]
a) 0
b) 1
c) -1
d) e

Hint: To solve the question, we have to apply inverse trigonometric functions properties and the concept of logarithms to simplify the given expression. For further solving apply L'Hospital's rule and formulae of differentiation to obtain the answer.

Complete step by step solution:
Let the given expression be equal to a

\[\Rightarrow a=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \dfrac{\pi }{2}-{{\tan }^{-1}}x \right)}^{{1}/{x}\;}}\]

We know that \[\dfrac{\pi }{2}-{{\tan }^{-1}}x={{\cot }^{-1}}x\]

By substituting the value, we get

\[a=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( {{\cot }^{-1}}x \right)}^{{1}/{x}\;}}\]

By applying log on both sides of equation we get,

\[\log a=\underset{x\to \infty }{\mathop{\lim }}\,\log {{\left( {{\cot }^{-1}}x \right)}^{{1}/{x}\;}}\]

We know the formula \[\log {{\left( x \right)}^{m}}=m\log x\]

By substituting the formula, we get

\[\log a=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{x}\log \left( {{\cot }^{-1}}x \right)\]
By applying L Hospital’s rule which states \[\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{f}^{1}}(x)}{{{g}^{1}}(x)}\] , we get

 \[\log a=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\dfrac{d\log \left( {{\cot }^{-1}}x \right)}{dx}}{\dfrac{dx}{dx}}\]

\[\log a=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\dfrac{d\log \left( {{\cot }^{-1}}x \right)}{dx}}{1}=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{d\log \left( {{\cot }^{-1}}x \right)}{dx}\]

We know the formula \[\dfrac{d\log f(x)}{dx}=\dfrac{1}{f(x)}\dfrac{df(x)}{dx}\]

Thus, we get

\[\log a=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{{{\cot }^{-1}}x}\dfrac{d\left( {{\cot }^{-1}}x \right)}{dx}\]

We know the formula \[\dfrac{d\left( {{\cot }^{-1}}x \right)}{dx}=\dfrac{-1}{1+{{x}^{2}}}\]

\[\log a=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{{{\cot }^{-1}}x}\left( \dfrac{-1}{1+{{x}^{2}}} \right)\]

\[\log a=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\left( \dfrac{-1}{1+{{x}^{2}}} \right)}{{{\cot }^{-1}}x}\]

By applying L'Hospital's rule, we get

\[\log a=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\dfrac{d\left( \dfrac{-1}{1+{{x}^{2}}} \right)}{dx}}{\dfrac{d\left( {{\cot }^{-1}}x \right)}{dx}}\]

We know the formula \[\dfrac{d\left( \dfrac{1}{f(x)} \right)}{dx}=\dfrac{-1}{{{f}^{2}}(x)}\dfrac{df(x)}{dx}\]

By applying the formula for numerator and denominator, we get

\[\log a=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\left( \dfrac{-(-1)}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right)\dfrac{d\left( 1+{{x}^{2}} \right)}{dx}}{\left( \dfrac{-1}{1+{{x}^{2}}} \right)}\]

\[oga=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\left( \dfrac{1}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right)\dfrac{d\left( {{x}^{2}} \right)}{dx}}{\left( \dfrac{-1}{1+{{x}^{2}}} \right)}\]
We know the formula \[\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}\]

\[\log a=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\left( \dfrac{1}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right)2x}{\left( \dfrac{-1}{1+{{x}^{2}}} \right)}\]

\[\log a=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\left( \dfrac{2x}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right)}{\left( \dfrac{-1}{1+{{x}^{2}}} \right)}\]

By eliminating the common terms in numerator and denominator of the fraction, we get
\[\log a=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\left( \dfrac{2x}{\left( 1+{{x}^{2}} \right)} \right)}{\left( -1 \right)}\]

\[\log a=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{-2x}{\left( 1+{{x}^{2}} \right)}\]
By rearranging the terms of the expression, we get

\[\log a=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{-2}{\dfrac{1}{x}\left( 1+{{x}^{2}} \right)}\]

\[\log a=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{-2}{\left( \dfrac{1}{x}+\dfrac{{{x}^{2}}}{x} \right)}\]

\[\log a=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{-2}{\left( \dfrac{1}{x}+x \right)}\]
We know that if \[x\to \infty \] then \[\dfrac{1}{x}\to 0\]

Thus, by substituting the values we get

\[\log a=\dfrac{-2}{\left( 0+\infty \right)}\]

\[\log a=\dfrac{-2}{\infty }\]

We know the value of \[\dfrac{1}{\infty }\] is equal to 0. Thus, by substituting the values we get

\[\log a=\dfrac{-2}{\infty }=0\]

\[\log a=0\]

We know that the value of \[\log x\] is equal to 0 when x = 1. Thus, by substituting the values we get

The value of a = 1.

Thus, we get the value of \[\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \dfrac{\pi }{2}-{{\tan }^{-1}}x \right)}^{{1}/{x}\;}}\]is equal to 1.

Hence, option (b) is the right choice.

Note: The possibility of mistake can be not applying the required formulae of algebra, inverse trigonometric functions and logarithms. The other possible mistake can be not applying L Hospital’s rule and not applying log which ease the procedure of solving.