Evaluate the given limit of the function: \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\cot }^{-1}}\left( \dfrac{1}{x} \right)}{x}\]
(a) Exists and is equal to one
(b) Does not exist as R.H.L is 1 and L.H.L. is -1
(c) Does not exist as R.H.L and L.H.L both are non-existent
(d) Does not exist as R.H.L exits but L.H.L does not
Answer
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Hint: Apply the limit directly and observe that we will get an indeterminate form. Use L’Hopital Rule and chain rule of composition of two functions to find the exact limit of the given function.
Complete step-by-step answer:
We are given the function \[\dfrac{{{\cot }^{-1}}\left( \dfrac{1}{x} \right)}{x}\]. We have to evaluate the limit \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\cot }^{-1}}\left( \dfrac{1}{x} \right)}{x}\]. We observe that if we simply apply the limit, we will get \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\cot }^{-1}}\left( \dfrac{1}{x} \right)}{x}=\dfrac{{{\cot }^{-1}}\left( \dfrac{1}{0} \right)}{0}=\dfrac{{{\cot }^{-1}}\left( \infty \right)}{0}=\dfrac{0}{0}\].
Hence, we will use L’Hopital Rule to evaluate the limit which states that if \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{0}{0}\] then we have \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}=\dfrac{f'\left( a \right)}{g'\left( a \right)}\].
Substituting \[f\left( x \right)={{\cot }^{-1}}\left( \dfrac{1}{x} \right),g\left( x \right)=x\] in the above equation, we have \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\cot }^{-1}}\left( \dfrac{1}{x} \right)}{x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)}{\dfrac{d}{dx}\left( x \right)}.....\left( 1 \right)\].
To find the value of \[\dfrac{d}{dx}\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)\], we will write \[y={{\cot }^{-1}}\left( \dfrac{1}{x} \right)\] as a composition of two functions \[y=u\left( v\left( x \right) \right)\] where \[u\left( x \right)={{\cot }^{-1}}x,v\left( x \right)=\dfrac{1}{x}\].
We will use chain rule of composition of two functions which states that if \[y=u\left( v\left( x \right) \right)\] then \[\dfrac{dy}{dx}=\dfrac{du\left( v\left( x \right) \right)}{dv\left( x \right)}\times \dfrac{dv\left( x \right)}{dx}\].
Substituting \[u\left( x \right)={{\cot }^{-1}}x,v\left( x \right)=\dfrac{1}{x}\] in the above formula, we have \[\dfrac{dy}{dx}=\dfrac{d\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)}{d\left( \dfrac{1}{x} \right)}\times \dfrac{d\left( \dfrac{1}{x} \right)}{dx}.....\left( 2 \right)\].
To find the value of \[\dfrac{d\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)}{d\left( \dfrac{1}{x} \right)}\], let’s assume \[t=\dfrac{1}{x}\].
Thus, we have \[\dfrac{d\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)}{d\left( \dfrac{1}{x} \right)}=\dfrac{d\left( {{\cot }^{-1}}t \right)}{dt}\].
We know that differentiation of any function of the form \[y={{\cot }^{-1}}\left( x \right)\] is \[\dfrac{d\left( {{\cot }^{-1}}x \right)}{dx}=\dfrac{-1}{1+{{x}^{2}}}\].
Thus, we have \[\dfrac{d\left( {{\cot }^{-1}}t \right)}{dt}=\dfrac{-1}{1+{{t}^{2}}}\].
So, we will get \[\dfrac{d\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)}{d\left( \dfrac{1}{x} \right)}=\dfrac{d\left( {{\cot }^{-1}}t \right)}{dt}=\dfrac{-1}{1+{{t}^{2}}}=\dfrac{-1}{1+\dfrac{1}{{{x}^{2}}}}\].
Simplifying the above equation, we get \[\dfrac{d\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)}{d\left( \dfrac{1}{x} \right)}=\dfrac{-1}{1+\dfrac{1}{{{x}^{2}}}}=\dfrac{-{{x}^{2}}}{1+{{x}^{2}}}.....\left( 3 \right)\].
To find the value of \[\dfrac{d\left( \dfrac{1}{x} \right)}{dx}\], substitute \[a=1,n=-1\] in the formula where if \[y=a{{x}^{n}}\] then \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Thus, we have \[\dfrac{d\left( \dfrac{1}{x} \right)}{dx}=\dfrac{-1}{{{x}^{2}}}.....\left( 4 \right)\].
Substituting equation \[\left( 3 \right)\] and \[\left( 4 \right)\] in equation \[\left( 2 \right)\], we get \[\dfrac{dy}{dx}=\dfrac{d\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)}{d\left( \dfrac{1}{x} \right)}\times \dfrac{d\left( \dfrac{1}{x} \right)}{dx}=\dfrac{-{{x}^{2}}}{1+{{x}^{2}}}\times \dfrac{-1}{{{x}^{2}}}=\dfrac{1}{1+{{x}^{2}}}.....\left( 5 \right)\].
To find the value of \[\dfrac{d}{dx}\left( x \right)\], substitute \[a=1,n=1\] in the formula where if \[y=a{{x}^{n}}\] then \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Thus, we have \[\dfrac{d}{dx}\left( x \right)=1.....\left( 6 \right)\].
Substituting the equation \[\left( 5 \right)\] and \[\left( 6 \right)\] in equation \[\left( 1 \right)\], we get \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\cot }^{-1}}\left( \dfrac{1}{x} \right)}{x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)}{\dfrac{d}{dx}\left( x \right)}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{1+{{x}^{2}}}=1\].
Hence, we observe that the limit exists and is equal to 1, which is option (a).
Note: It’s very necessary to use the L'Hopital Rule to find the limit of the given function. We won’t get the correct answer by directly substituting the limit. An indeterminate form is an expression involving two functions whose limit can’t be determined solely from the limits of the individual functions. Indeterminate forms of the functions include \[\dfrac{0}{0},\dfrac{\infty }{\infty },0\times \infty ,\infty -\infty ,{{0}^{0}},{{1}^{\infty }},{{\infty }^{0}}\].
Complete step-by-step answer:
We are given the function \[\dfrac{{{\cot }^{-1}}\left( \dfrac{1}{x} \right)}{x}\]. We have to evaluate the limit \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\cot }^{-1}}\left( \dfrac{1}{x} \right)}{x}\]. We observe that if we simply apply the limit, we will get \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\cot }^{-1}}\left( \dfrac{1}{x} \right)}{x}=\dfrac{{{\cot }^{-1}}\left( \dfrac{1}{0} \right)}{0}=\dfrac{{{\cot }^{-1}}\left( \infty \right)}{0}=\dfrac{0}{0}\].
Hence, we will use L’Hopital Rule to evaluate the limit which states that if \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{0}{0}\] then we have \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}=\dfrac{f'\left( a \right)}{g'\left( a \right)}\].
Substituting \[f\left( x \right)={{\cot }^{-1}}\left( \dfrac{1}{x} \right),g\left( x \right)=x\] in the above equation, we have \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\cot }^{-1}}\left( \dfrac{1}{x} \right)}{x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)}{\dfrac{d}{dx}\left( x \right)}.....\left( 1 \right)\].
To find the value of \[\dfrac{d}{dx}\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)\], we will write \[y={{\cot }^{-1}}\left( \dfrac{1}{x} \right)\] as a composition of two functions \[y=u\left( v\left( x \right) \right)\] where \[u\left( x \right)={{\cot }^{-1}}x,v\left( x \right)=\dfrac{1}{x}\].
We will use chain rule of composition of two functions which states that if \[y=u\left( v\left( x \right) \right)\] then \[\dfrac{dy}{dx}=\dfrac{du\left( v\left( x \right) \right)}{dv\left( x \right)}\times \dfrac{dv\left( x \right)}{dx}\].
Substituting \[u\left( x \right)={{\cot }^{-1}}x,v\left( x \right)=\dfrac{1}{x}\] in the above formula, we have \[\dfrac{dy}{dx}=\dfrac{d\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)}{d\left( \dfrac{1}{x} \right)}\times \dfrac{d\left( \dfrac{1}{x} \right)}{dx}.....\left( 2 \right)\].
To find the value of \[\dfrac{d\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)}{d\left( \dfrac{1}{x} \right)}\], let’s assume \[t=\dfrac{1}{x}\].
Thus, we have \[\dfrac{d\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)}{d\left( \dfrac{1}{x} \right)}=\dfrac{d\left( {{\cot }^{-1}}t \right)}{dt}\].
We know that differentiation of any function of the form \[y={{\cot }^{-1}}\left( x \right)\] is \[\dfrac{d\left( {{\cot }^{-1}}x \right)}{dx}=\dfrac{-1}{1+{{x}^{2}}}\].
Thus, we have \[\dfrac{d\left( {{\cot }^{-1}}t \right)}{dt}=\dfrac{-1}{1+{{t}^{2}}}\].
So, we will get \[\dfrac{d\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)}{d\left( \dfrac{1}{x} \right)}=\dfrac{d\left( {{\cot }^{-1}}t \right)}{dt}=\dfrac{-1}{1+{{t}^{2}}}=\dfrac{-1}{1+\dfrac{1}{{{x}^{2}}}}\].
Simplifying the above equation, we get \[\dfrac{d\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)}{d\left( \dfrac{1}{x} \right)}=\dfrac{-1}{1+\dfrac{1}{{{x}^{2}}}}=\dfrac{-{{x}^{2}}}{1+{{x}^{2}}}.....\left( 3 \right)\].
To find the value of \[\dfrac{d\left( \dfrac{1}{x} \right)}{dx}\], substitute \[a=1,n=-1\] in the formula where if \[y=a{{x}^{n}}\] then \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Thus, we have \[\dfrac{d\left( \dfrac{1}{x} \right)}{dx}=\dfrac{-1}{{{x}^{2}}}.....\left( 4 \right)\].
Substituting equation \[\left( 3 \right)\] and \[\left( 4 \right)\] in equation \[\left( 2 \right)\], we get \[\dfrac{dy}{dx}=\dfrac{d\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)}{d\left( \dfrac{1}{x} \right)}\times \dfrac{d\left( \dfrac{1}{x} \right)}{dx}=\dfrac{-{{x}^{2}}}{1+{{x}^{2}}}\times \dfrac{-1}{{{x}^{2}}}=\dfrac{1}{1+{{x}^{2}}}.....\left( 5 \right)\].
To find the value of \[\dfrac{d}{dx}\left( x \right)\], substitute \[a=1,n=1\] in the formula where if \[y=a{{x}^{n}}\] then \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Thus, we have \[\dfrac{d}{dx}\left( x \right)=1.....\left( 6 \right)\].
Substituting the equation \[\left( 5 \right)\] and \[\left( 6 \right)\] in equation \[\left( 1 \right)\], we get \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\cot }^{-1}}\left( \dfrac{1}{x} \right)}{x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( {{\cot }^{-1}}\left( \dfrac{1}{x} \right) \right)}{\dfrac{d}{dx}\left( x \right)}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{1+{{x}^{2}}}=1\].
Hence, we observe that the limit exists and is equal to 1, which is option (a).
Note: It’s very necessary to use the L'Hopital Rule to find the limit of the given function. We won’t get the correct answer by directly substituting the limit. An indeterminate form is an expression involving two functions whose limit can’t be determined solely from the limits of the individual functions. Indeterminate forms of the functions include \[\dfrac{0}{0},\dfrac{\infty }{\infty },0\times \infty ,\infty -\infty ,{{0}^{0}},{{1}^{\infty }},{{\infty }^{0}}\].
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