Answer
Verified
425.4k+ views
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.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE