Evaluate \[\underset{x\to \sqrt{6}}{\mathop{\lim }}\,\dfrac{\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{{{x}^{2}}-6}\].
Answer
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Hint: Use L’Hopital Rule which states that if \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{f\left( a \right)}{g\left( a \right)}=\dfrac{0}{0}\], we will evaluate the limit by \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}=\dfrac{f'\left( a \right)}{g'\left( a \right)}\], to find the limit of the given function.
Complete step-by-step answer:
To evaluate the limit, we will find the left and right hand side of the limit by substituting the given limit in the equation of function.
Thus, by applying right side of the limit, we have \[\underset{x\to {{\sqrt{6}}^{+}}}{\mathop{\lim }}\,\dfrac{\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{{{x}^{2}}-6}=\dfrac{\left( \sqrt{5+2\sqrt{6}}-\sqrt{3}+\sqrt{2} \right)}{{{\left( \sqrt{6} \right)}^{2}}-6}=\dfrac{\left( \sqrt{{{\left( \sqrt{3}+\sqrt{2} \right)}^{2}}-\sqrt{3}+\sqrt{2}} \right)}{6-6}=\infty \].
When we apply left side of the limit, we get \[\underset{x\to {{\sqrt{6}}^{-}}}{\mathop{\lim }}\,\dfrac{\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{{{x}^{2}}-6}=\dfrac{\left( \sqrt{5+2\sqrt{6}}-\sqrt{3}+\sqrt{2} \right)}{{{\left( \sqrt{6} \right)}^{2}}-6}=\dfrac{\left( \sqrt{\sqrt{{{\left( \sqrt{3}+\sqrt{2} \right)}^{2}}}-\sqrt{3}+\sqrt{2}} \right)}{6-6}=\dfrac{\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}}{-0}=-\infty \]
As \[\underset{x\to {{\sqrt{6}}^{+}}}{\mathop{\lim }}\,\dfrac{\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{{{x}^{2}}-6}=\infty \ne \underset{x\to {{\sqrt{6}}^{-}}}{\mathop{\lim }}\,\dfrac{\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{{{x}^{2}}-6}=-\infty \], we will use L’Hopital Rule to find the limit of the function which states that if \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{f\left( a \right)}{g\left( a \right)}=\dfrac{0}{0}\], and then we will find the limit by \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}=\dfrac{f'\left( a \right)}{g'\left( a \right)}\].
Substituting \[a=\sqrt{6},f\left( x \right)=\sqrt{5+2x}-\sqrt{3}+\sqrt{2},g\left( x \right)={{x}^{2}}-6\] in the above equation, we have \[\underset{x\to \sqrt{6}}{\mathop{\lim }}\,\dfrac{\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{{{x}^{2}}-6}=\underset{x\to \sqrt{6}}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{\dfrac{d}{dx}({{x}^{2}}-6)}.....\left( 1 \right)\].
To find the value of \[\dfrac{d}{dx}\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)\], we will write \[f\left( x \right)=\sqrt{5+2x}-\sqrt{3}+\sqrt{2}\] as a composition of two functions \[f\left( x \right)=a\left( x \right)+b\left( x \right)\] where \[a\left( x \right)=\sqrt{5+2x},b\left( x \right)=\sqrt{2}-\sqrt{3}\].
We will use sum rule of differentiation of two functions which states that if \[y=a\left( x \right)+b\left( x \right)\] then \[\dfrac{dy}{dx}=\dfrac{d}{dx}a\left( x \right)+\dfrac{d}{dx}b\left( x \right)\].
Substituting \[a\left( x \right)=\sqrt{5+2x},b\left( x \right)=\sqrt{2}-\sqrt{3}\] in the above equation, we have \[\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \sqrt{5+2x} \right)+\dfrac{d}{dx}\left( \sqrt{2}-\sqrt{3} \right).....\left( 2 \right)\].
To find the value of \[\dfrac{d}{dx}a\left( x \right)=\dfrac{d}{dx}\left( \sqrt{5+2x} \right)\], we will write \[a\left( x \right)\] as a composition of two functions \[a\left( x \right)=u\left( v\left( x \right) \right)\] where \[u\left( x \right)=\sqrt{x},v\left( x \right)=5+2x\].
We will use chain rule of composition of differentiation of two functions which states that if \[y=a\left( x \right)=u\left( v\left( x \right) \right)\] then \[\dfrac{dy}{dx}=\dfrac{du\left( v\left( x \right) \right)}{dv\left( x \right)}\times \dfrac{dv\left( x \right)}{dx}\].
Substituting \[u\left( x \right)=\sqrt{x},v\left( x \right)=5+2x\] in the above equation, we have \[\dfrac{dy}{dx}=\dfrac{du\left( v\left( x \right) \right)}{dv\left( x \right)}\times \dfrac{dv\left( x \right)}{dx}=\dfrac{d\left( \sqrt{5+2x} \right)}{d\left( 5+2x \right)}\times \dfrac{d}{dx}\left( 5+2x \right).....\left( 3 \right)\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=2,n=1,b=5\] in the above equation, we have \[\dfrac{d}{dx}\left( 5+2x \right)=2.....\left( 4 \right)\].
To find the value of \[\dfrac{d\left( \sqrt{5+2x} \right)}{d\left( 5+2x \right)}\], let’s assume \[t=5+2x\].
Thus, we have \[\dfrac{d\left( \sqrt{5+2x} \right)}{d\left( 5+2x \right)}=\dfrac{d\left( \sqrt{t} \right)}{dt}\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=1,n=\dfrac{1}{2},b=0\] in the above equation, we have \[\dfrac{d\left( \sqrt{t} \right)}{dt}=\dfrac{1}{2\sqrt{t}}\].
Thus, we get \[\dfrac{d\left( \sqrt{5+2x} \right)}{d\left( 5+2x \right)}=\dfrac{d\left( \sqrt{t} \right)}{dt}=\dfrac{1}{2\sqrt{t}}=\dfrac{1}{2\sqrt{5+2x}}.....\left( 5 \right)\].
Substituting equation \[\left( 4 \right)\] and \[\left( 5 \right)\] in equation \[\left( 3 \right)\], we get \[\dfrac{d}{dx}a\left( x \right)=\dfrac{d\left( \sqrt{5+2x} \right)}{d\left( 5+2x \right)}\times \dfrac{d}{dx}\left( 5+2x \right)=\dfrac{1}{2\sqrt{5+2x}}\times 2=\dfrac{1}{\sqrt{5+2x}}.....\left( 6 \right)\].
We know that differentiation of a constant is zero. Thus, we have \[\dfrac{d}{dx}b\left( x \right)=0.....\left( 7 \right)\].
Substituting equation \[\left( 6 \right)\] and \[\left( 7 \right)\] in equation \[\left( 2 \right)\], we get \[\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( \sqrt{5+2x} \right)+\dfrac{d}{dx}\left( \sqrt{2}-\sqrt{3} \right)=\dfrac{1}{\sqrt{5+2x}}.....\left( 8 \right)\].
To find the value of \[\dfrac{d}{dx}({{x}^{2}}-6)\], we will substitute \[a=1,n=2,b=-6\] in the equation where differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Thus, we have \[\dfrac{d}{dx}g\left( x \right)=2x.....\left( 9 \right)\].
Substituting equation \[\left( 8 \right)\] and \[\left( 9 \right)\] in equation \[\left( 1 \right)\], we get \[\underset{x\to \sqrt{6}}{\mathop{\lim }}\,\dfrac{\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{{{x}^{2}}-6}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{\dfrac{d}{dx}({{x}^{2}}-6)}=\underset{x\to \sqrt{6}}{\mathop{\lim }}\,\dfrac{\dfrac{1}{\sqrt{5+2x}}}{2x}\].
Solving the above equation, we get \[\underset{x\to \sqrt{6}}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{\dfrac{d}{dx}({{x}^{2}}-6)}=\underset{x\to \sqrt{6}}{\mathop{\lim }}\,\dfrac{\dfrac{1}{\sqrt{5+2x}}}{2x}=\underset{x\to \sqrt{6}}{\mathop{\lim }}\,\dfrac{1}{2x\sqrt{5+2x}}=\dfrac{1}{2\sqrt{6}\sqrt{5+2\sqrt{6}}}\].
Thus, we have \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{{{x}^{2}}-6}=\dfrac{1}{2\sqrt{6}\sqrt{5+2\sqrt{6}}}\].
Note: We won’t get the correct answer without the use of L’Hopital Rule and using the identity \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]. If we simply substitute the values in the given equation, we will get an incorrect answer. Also, one must carefully differentiate the functions in the numerator and denominator.
Complete step-by-step answer:
To evaluate the limit, we will find the left and right hand side of the limit by substituting the given limit in the equation of function.
Thus, by applying right side of the limit, we have \[\underset{x\to {{\sqrt{6}}^{+}}}{\mathop{\lim }}\,\dfrac{\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{{{x}^{2}}-6}=\dfrac{\left( \sqrt{5+2\sqrt{6}}-\sqrt{3}+\sqrt{2} \right)}{{{\left( \sqrt{6} \right)}^{2}}-6}=\dfrac{\left( \sqrt{{{\left( \sqrt{3}+\sqrt{2} \right)}^{2}}-\sqrt{3}+\sqrt{2}} \right)}{6-6}=\infty \].
When we apply left side of the limit, we get \[\underset{x\to {{\sqrt{6}}^{-}}}{\mathop{\lim }}\,\dfrac{\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{{{x}^{2}}-6}=\dfrac{\left( \sqrt{5+2\sqrt{6}}-\sqrt{3}+\sqrt{2} \right)}{{{\left( \sqrt{6} \right)}^{2}}-6}=\dfrac{\left( \sqrt{\sqrt{{{\left( \sqrt{3}+\sqrt{2} \right)}^{2}}}-\sqrt{3}+\sqrt{2}} \right)}{6-6}=\dfrac{\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}}{-0}=-\infty \]
As \[\underset{x\to {{\sqrt{6}}^{+}}}{\mathop{\lim }}\,\dfrac{\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{{{x}^{2}}-6}=\infty \ne \underset{x\to {{\sqrt{6}}^{-}}}{\mathop{\lim }}\,\dfrac{\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{{{x}^{2}}-6}=-\infty \], we will use L’Hopital Rule to find the limit of the function which states that if \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{f\left( a \right)}{g\left( a \right)}=\dfrac{0}{0}\], and then we will find the limit by \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}=\dfrac{f'\left( a \right)}{g'\left( a \right)}\].
Substituting \[a=\sqrt{6},f\left( x \right)=\sqrt{5+2x}-\sqrt{3}+\sqrt{2},g\left( x \right)={{x}^{2}}-6\] in the above equation, we have \[\underset{x\to \sqrt{6}}{\mathop{\lim }}\,\dfrac{\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{{{x}^{2}}-6}=\underset{x\to \sqrt{6}}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{\dfrac{d}{dx}({{x}^{2}}-6)}.....\left( 1 \right)\].
To find the value of \[\dfrac{d}{dx}\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)\], we will write \[f\left( x \right)=\sqrt{5+2x}-\sqrt{3}+\sqrt{2}\] as a composition of two functions \[f\left( x \right)=a\left( x \right)+b\left( x \right)\] where \[a\left( x \right)=\sqrt{5+2x},b\left( x \right)=\sqrt{2}-\sqrt{3}\].
We will use sum rule of differentiation of two functions which states that if \[y=a\left( x \right)+b\left( x \right)\] then \[\dfrac{dy}{dx}=\dfrac{d}{dx}a\left( x \right)+\dfrac{d}{dx}b\left( x \right)\].
Substituting \[a\left( x \right)=\sqrt{5+2x},b\left( x \right)=\sqrt{2}-\sqrt{3}\] in the above equation, we have \[\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \sqrt{5+2x} \right)+\dfrac{d}{dx}\left( \sqrt{2}-\sqrt{3} \right).....\left( 2 \right)\].
To find the value of \[\dfrac{d}{dx}a\left( x \right)=\dfrac{d}{dx}\left( \sqrt{5+2x} \right)\], we will write \[a\left( x \right)\] as a composition of two functions \[a\left( x \right)=u\left( v\left( x \right) \right)\] where \[u\left( x \right)=\sqrt{x},v\left( x \right)=5+2x\].
We will use chain rule of composition of differentiation of two functions which states that if \[y=a\left( x \right)=u\left( v\left( x \right) \right)\] then \[\dfrac{dy}{dx}=\dfrac{du\left( v\left( x \right) \right)}{dv\left( x \right)}\times \dfrac{dv\left( x \right)}{dx}\].
Substituting \[u\left( x \right)=\sqrt{x},v\left( x \right)=5+2x\] in the above equation, we have \[\dfrac{dy}{dx}=\dfrac{du\left( v\left( x \right) \right)}{dv\left( x \right)}\times \dfrac{dv\left( x \right)}{dx}=\dfrac{d\left( \sqrt{5+2x} \right)}{d\left( 5+2x \right)}\times \dfrac{d}{dx}\left( 5+2x \right).....\left( 3 \right)\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=2,n=1,b=5\] in the above equation, we have \[\dfrac{d}{dx}\left( 5+2x \right)=2.....\left( 4 \right)\].
To find the value of \[\dfrac{d\left( \sqrt{5+2x} \right)}{d\left( 5+2x \right)}\], let’s assume \[t=5+2x\].
Thus, we have \[\dfrac{d\left( \sqrt{5+2x} \right)}{d\left( 5+2x \right)}=\dfrac{d\left( \sqrt{t} \right)}{dt}\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=1,n=\dfrac{1}{2},b=0\] in the above equation, we have \[\dfrac{d\left( \sqrt{t} \right)}{dt}=\dfrac{1}{2\sqrt{t}}\].
Thus, we get \[\dfrac{d\left( \sqrt{5+2x} \right)}{d\left( 5+2x \right)}=\dfrac{d\left( \sqrt{t} \right)}{dt}=\dfrac{1}{2\sqrt{t}}=\dfrac{1}{2\sqrt{5+2x}}.....\left( 5 \right)\].
Substituting equation \[\left( 4 \right)\] and \[\left( 5 \right)\] in equation \[\left( 3 \right)\], we get \[\dfrac{d}{dx}a\left( x \right)=\dfrac{d\left( \sqrt{5+2x} \right)}{d\left( 5+2x \right)}\times \dfrac{d}{dx}\left( 5+2x \right)=\dfrac{1}{2\sqrt{5+2x}}\times 2=\dfrac{1}{\sqrt{5+2x}}.....\left( 6 \right)\].
We know that differentiation of a constant is zero. Thus, we have \[\dfrac{d}{dx}b\left( x \right)=0.....\left( 7 \right)\].
Substituting equation \[\left( 6 \right)\] and \[\left( 7 \right)\] in equation \[\left( 2 \right)\], we get \[\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( \sqrt{5+2x} \right)+\dfrac{d}{dx}\left( \sqrt{2}-\sqrt{3} \right)=\dfrac{1}{\sqrt{5+2x}}.....\left( 8 \right)\].
To find the value of \[\dfrac{d}{dx}({{x}^{2}}-6)\], we will substitute \[a=1,n=2,b=-6\] in the equation where differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Thus, we have \[\dfrac{d}{dx}g\left( x \right)=2x.....\left( 9 \right)\].
Substituting equation \[\left( 8 \right)\] and \[\left( 9 \right)\] in equation \[\left( 1 \right)\], we get \[\underset{x\to \sqrt{6}}{\mathop{\lim }}\,\dfrac{\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{{{x}^{2}}-6}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{\dfrac{d}{dx}({{x}^{2}}-6)}=\underset{x\to \sqrt{6}}{\mathop{\lim }}\,\dfrac{\dfrac{1}{\sqrt{5+2x}}}{2x}\].
Solving the above equation, we get \[\underset{x\to \sqrt{6}}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{\dfrac{d}{dx}({{x}^{2}}-6)}=\underset{x\to \sqrt{6}}{\mathop{\lim }}\,\dfrac{\dfrac{1}{\sqrt{5+2x}}}{2x}=\underset{x\to \sqrt{6}}{\mathop{\lim }}\,\dfrac{1}{2x\sqrt{5+2x}}=\dfrac{1}{2\sqrt{6}\sqrt{5+2\sqrt{6}}}\].
Thus, we have \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( \sqrt{5+2x}-\sqrt{3}+\sqrt{2} \right)}{{{x}^{2}}-6}=\dfrac{1}{2\sqrt{6}\sqrt{5+2\sqrt{6}}}\].
Note: We won’t get the correct answer without the use of L’Hopital Rule and using the identity \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]. If we simply substitute the values in the given equation, we will get an incorrect answer. Also, one must carefully differentiate the functions in the numerator and denominator.
Last updated date: 18th Sep 2023
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