# If f’’(x) be continuous at x=0.

If, $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{2f(x)-3af(2x)+bf(8x)}{{{\sin }^{2}}x}$ exists and $f(0)\ne 0,$ $f'(0)\ne 0,$

Then the value of $\dfrac{3a}{b}$___________________.

Last updated date: 19th Mar 2023

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Answer

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Hint: To solve the problem we can directly put the limits and see that with which indeterminate form it matches and then we can solve it accordingly by equating it with that form.

Firstly we will write the given values,

$f(0)\ne 0,$$f'(0)\ne 0,$……………………………………….. (1)

Now we will write the given expression and name it as L,

$L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{2f(x)-3af(2x)+bf(8x)}{{{\sin }^{2}}x}$………………………………… (2)

As we don’t know what to do next, we will put the limits directly to see what happens with the above equation,

$\therefore L=\dfrac{2f(0)-3af(2\times 0)+bf(8\times 0)}{{{\sin }^{2}}\left( 0 \right)}$

\[\therefore L=\dfrac{2f(0)-3af(0)+bf(0)}{{{\sin }^{2}}0}\]

As we all know the value of sin 0 is zero, therefore we will get,

\[\therefore L=\dfrac{2f(0)-3af(0)+bf(0)}{0}\]

As it is mentioned that the limit ‘exists’ and if the limit exists then the value of the above equation should match with one of the indeterminate forms so that we can find the limit using L-Hospitals Rule.

As the denominator of the given limit is becoming zero therefore we can say that it is of the $\dfrac{0}{0}$ form which is a type of indeterminate form.

Therefore we can conclude that both the numerator and denominator are zero. Therefore we will get,

\[\therefore 2f(0)-3af(0)+bf(0)=0\]

Taking f(0) we will get,

\[\therefore f(0)\times \left( 2-3a+b \right)=0\]

As it is given in (1) that f(0) is not equal to zero, therefore we will get,

\[\left( 2-3a+b \right)=0\]

\[\therefore 2+b=3a\]………………………………………………. (3)

Now as we know that the L is giving the indeterminate form therefore we can use the L-Hospital’s Rule which is given below,

L-Hospital’s Rule:

\[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}f(x)}{\dfrac{d}{dx}g(x)}\]

By using above rule we will write the equation (2) as shown below,

\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left[ 2f(x)-3af(2x)+bf(8x) \right]}{\dfrac{d}{dx}\left( {{\sin }^{2}}x \right)}\]

To solve further we should know the formulae given below,

Formulae:

\[\begin{align}

& 1.\dfrac{d}{dx}f(x)=f'(x) \\

& 2.\dfrac{d}{dx}\sin x=\cos x \\

& 3.\dfrac{d}{dx}f{{(x)}^{2}}=2f(x)\times f'(x) \\

\end{align}\]

By using Formulae (1) in numerator of ‘L’ we will get,

\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left[ 2f'(x)-3af'(2x)\dfrac{d}{dx}2x+bf'(8x)\dfrac{d}{dx}8x \right]}{\dfrac{d}{dx}\left( {{\sin }^{2}}x \right)}\]

By using formula (3) in denominator we will get,

\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left[ 2f'(x)-3af'(2x)\times 2+bf'(8x)\times 8 \right]}{\left( 2\sin x \right)\dfrac{d}{dx}\left( \sin x \right)}\]

By using the formula (2) in the denominator we can get,

\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left[ 2f'(x)-6af'(2x)+8bf'(8x) \right]}{2\sin x\times \cos x}\]

To proceed further we should the formula given below,

Formula:

\[\sin 2x=2\times \sin x\times \cos x\]

By using above formula we can write L as,

\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left[ 2f'(x)-6af'(2x)+8bf'(8x) \right]}{\sin 2x}\]

Now if we put the limits directly we will get,

\[L=\dfrac{\left[ 2f'(0)-6af'(2\times 0)+8bf'(8\times 0) \right]}{\sin \left( 2\times 0 \right)}\]

\[L=\dfrac{\left[ 2f'(0)-6af'(0)+8bf'(0) \right]}{\sin \left( 0 \right)}\]

As we all know the value of sin 0 is zero,

\[\therefore L=\dfrac{\left[ 2f'(0)-6af'(0)+8bf'(0) \right]}{0}\]

Again we are getting the condition of indeterminate form and we can easily say that it’s a \[\dfrac{0}{0}\] form and to make the above equation in this form, the Numerator should have to be Zero.

\[\therefore 2f'(0)-6af'(0)+8bf'(0)=0\]

Taking f’(0) common we will get,

\[\therefore f'(x)\left( 2-6a+8b \right)=0\]

If we see the equation (1) we will come to know that the value of f’(x) is not equal to Zero.

\[\therefore \left( 2-2\left( 3a \right)+8b \right)=0\]………………………………….. (4)

Now we will put the value of equation (3) in above equation we will get,

\[\therefore \left[ 2-2\left( b+2 \right)+8b \right]=0\]

\[\therefore \left[ 2-2b-4+8b \right]=0\]

\[\therefore 6b-2=0\]

\[\therefore b=\dfrac{1}{3}\]………………………………………………………….. (5)

Now put the value of ‘b’ in equation (4),

\[\therefore \left( 2-2\left( 3a \right)+8\left( \dfrac{1}{3} \right) \right)=0\]

\[\therefore 2-6a+\dfrac{8}{3}=0\]

\[\therefore 2+\dfrac{8}{3}=6a\]

\[\therefore \dfrac{6+8}{3}=6a\]

\[\therefore \dfrac{14}{3\times 6}=a\]

\[\therefore \dfrac{7}{3\times 3}=a\]

\[\therefore a=\dfrac{7}{9}\]………………………………………………………….. (6)

Now consider, \[\dfrac{3a}{b}\] and put the values of (5) and (6) in it, then we will get,

\[\therefore \dfrac{3a}{b}=\dfrac{3\times \dfrac{7}{9}}{\dfrac{1}{3}}\]

\[\therefore \dfrac{3a}{b}=\dfrac{\dfrac{7}{3}}{\dfrac{1}{3}}\]

\[\therefore \dfrac{3a}{b}=\dfrac{7}{3}\times \dfrac{3}{1}\]

\[\therefore \dfrac{3a}{b}=7\]

Therefore the value of \[\dfrac{3a}{b}\] is 7.

Note: While taking the derivative of a given expression do remember to use chain rule in f(x) also, If we won’t use the chain rule then we will not be able to get any answer.

Chain Rule:

\[.\dfrac{d}{dx}f{{(x)}^{2}}=2f(x)\times \dfrac{d}{dx}f(x)\]

Firstly we will write the given values,

$f(0)\ne 0,$$f'(0)\ne 0,$……………………………………….. (1)

Now we will write the given expression and name it as L,

$L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{2f(x)-3af(2x)+bf(8x)}{{{\sin }^{2}}x}$………………………………… (2)

As we don’t know what to do next, we will put the limits directly to see what happens with the above equation,

$\therefore L=\dfrac{2f(0)-3af(2\times 0)+bf(8\times 0)}{{{\sin }^{2}}\left( 0 \right)}$

\[\therefore L=\dfrac{2f(0)-3af(0)+bf(0)}{{{\sin }^{2}}0}\]

As we all know the value of sin 0 is zero, therefore we will get,

\[\therefore L=\dfrac{2f(0)-3af(0)+bf(0)}{0}\]

As it is mentioned that the limit ‘exists’ and if the limit exists then the value of the above equation should match with one of the indeterminate forms so that we can find the limit using L-Hospitals Rule.

As the denominator of the given limit is becoming zero therefore we can say that it is of the $\dfrac{0}{0}$ form which is a type of indeterminate form.

Therefore we can conclude that both the numerator and denominator are zero. Therefore we will get,

\[\therefore 2f(0)-3af(0)+bf(0)=0\]

Taking f(0) we will get,

\[\therefore f(0)\times \left( 2-3a+b \right)=0\]

As it is given in (1) that f(0) is not equal to zero, therefore we will get,

\[\left( 2-3a+b \right)=0\]

\[\therefore 2+b=3a\]………………………………………………. (3)

Now as we know that the L is giving the indeterminate form therefore we can use the L-Hospital’s Rule which is given below,

L-Hospital’s Rule:

\[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}f(x)}{\dfrac{d}{dx}g(x)}\]

By using above rule we will write the equation (2) as shown below,

\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left[ 2f(x)-3af(2x)+bf(8x) \right]}{\dfrac{d}{dx}\left( {{\sin }^{2}}x \right)}\]

To solve further we should know the formulae given below,

Formulae:

\[\begin{align}

& 1.\dfrac{d}{dx}f(x)=f'(x) \\

& 2.\dfrac{d}{dx}\sin x=\cos x \\

& 3.\dfrac{d}{dx}f{{(x)}^{2}}=2f(x)\times f'(x) \\

\end{align}\]

By using Formulae (1) in numerator of ‘L’ we will get,

\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left[ 2f'(x)-3af'(2x)\dfrac{d}{dx}2x+bf'(8x)\dfrac{d}{dx}8x \right]}{\dfrac{d}{dx}\left( {{\sin }^{2}}x \right)}\]

By using formula (3) in denominator we will get,

\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left[ 2f'(x)-3af'(2x)\times 2+bf'(8x)\times 8 \right]}{\left( 2\sin x \right)\dfrac{d}{dx}\left( \sin x \right)}\]

By using the formula (2) in the denominator we can get,

\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left[ 2f'(x)-6af'(2x)+8bf'(8x) \right]}{2\sin x\times \cos x}\]

To proceed further we should the formula given below,

Formula:

\[\sin 2x=2\times \sin x\times \cos x\]

By using above formula we can write L as,

\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left[ 2f'(x)-6af'(2x)+8bf'(8x) \right]}{\sin 2x}\]

Now if we put the limits directly we will get,

\[L=\dfrac{\left[ 2f'(0)-6af'(2\times 0)+8bf'(8\times 0) \right]}{\sin \left( 2\times 0 \right)}\]

\[L=\dfrac{\left[ 2f'(0)-6af'(0)+8bf'(0) \right]}{\sin \left( 0 \right)}\]

As we all know the value of sin 0 is zero,

\[\therefore L=\dfrac{\left[ 2f'(0)-6af'(0)+8bf'(0) \right]}{0}\]

Again we are getting the condition of indeterminate form and we can easily say that it’s a \[\dfrac{0}{0}\] form and to make the above equation in this form, the Numerator should have to be Zero.

\[\therefore 2f'(0)-6af'(0)+8bf'(0)=0\]

Taking f’(0) common we will get,

\[\therefore f'(x)\left( 2-6a+8b \right)=0\]

If we see the equation (1) we will come to know that the value of f’(x) is not equal to Zero.

\[\therefore \left( 2-2\left( 3a \right)+8b \right)=0\]………………………………….. (4)

Now we will put the value of equation (3) in above equation we will get,

\[\therefore \left[ 2-2\left( b+2 \right)+8b \right]=0\]

\[\therefore \left[ 2-2b-4+8b \right]=0\]

\[\therefore 6b-2=0\]

\[\therefore b=\dfrac{1}{3}\]………………………………………………………….. (5)

Now put the value of ‘b’ in equation (4),

\[\therefore \left( 2-2\left( 3a \right)+8\left( \dfrac{1}{3} \right) \right)=0\]

\[\therefore 2-6a+\dfrac{8}{3}=0\]

\[\therefore 2+\dfrac{8}{3}=6a\]

\[\therefore \dfrac{6+8}{3}=6a\]

\[\therefore \dfrac{14}{3\times 6}=a\]

\[\therefore \dfrac{7}{3\times 3}=a\]

\[\therefore a=\dfrac{7}{9}\]………………………………………………………….. (6)

Now consider, \[\dfrac{3a}{b}\] and put the values of (5) and (6) in it, then we will get,

\[\therefore \dfrac{3a}{b}=\dfrac{3\times \dfrac{7}{9}}{\dfrac{1}{3}}\]

\[\therefore \dfrac{3a}{b}=\dfrac{\dfrac{7}{3}}{\dfrac{1}{3}}\]

\[\therefore \dfrac{3a}{b}=\dfrac{7}{3}\times \dfrac{3}{1}\]

\[\therefore \dfrac{3a}{b}=7\]

Therefore the value of \[\dfrac{3a}{b}\] is 7.

Note: While taking the derivative of a given expression do remember to use chain rule in f(x) also, If we won’t use the chain rule then we will not be able to get any answer.

Chain Rule:

\[.\dfrac{d}{dx}f{{(x)}^{2}}=2f(x)\times \dfrac{d}{dx}f(x)\]

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