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Question: How to derive this equation? Find a. \[\displaystyle \lim_{x \to 0}\left( \dfrac{\int_{0}^{x}{\dfr...

How to derive this equation? Find a.
lim⁡x→0(∫0xt2a+2t5dtbx−esin⁡x)=1π\displaystyle \lim_{x \to 0}\left( \dfrac{\int_{0}^{x}{\dfrac{{{t}^{2}}}{\sqrt{a+2{{t}^{5}}}}dt}}{bx-e\sin x} \right)=\dfrac{1}{\pi }

Explanation

Solution

For deriving the given equation, we need to consider the LHS, which is a limit of the form 00\dfrac{0}{0} so that we can apply the L-hospital’s rule. Therefore, we will differentiate both the numerator and the denominator. For differentiating the numerator, we will use the derivative of the definite integral given by ddx(∫k1(x)k2(x)f(t)dt)=d(k2(x))dxf(k2(x))−d(k1(x))dxf(k1(x))\dfrac{d}{dx}\left( \int_{{{k}_{1}}\left( x \right)}^{{{k}_{2}}\left( x \right)}{f\left( t \right)dt} \right)=\dfrac{d\left( {{k}_{2}}\left( x \right) \right)}{dx}f\left( {{k}_{2}}\left( x \right) \right)-\dfrac{d\left( {{k}_{1}}\left( x \right) \right)}{dx}f\left( {{k}_{1}}\left( x \right) \right). Then, we will obtain the value of b, with the help of which the limit will be obtained in terms of a. On equating the value of the limit obtained to the RHS, equal to 1π\dfrac{1}{\pi }, we will finally obtain the value of a, and hence the final answer.

Complete step by step solution:
The equation to be proved is given in the question as
⇒lim⁡x→0(∫0xt2a+2t5dtbx−esin⁡x)=1π\Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{\int_{0}^{x}{\dfrac{{{t}^{2}}}{\sqrt{a+2{{t}^{5}}}}dt}}{bx-e\sin x} \right)=\dfrac{1}{\pi }
Considering the LHS of the above equation, we have
⇒LHS=lim⁡x→0(∫0xt2a+2t5dtbx−esin⁡x).......(i)\Rightarrow LHS=\displaystyle \lim_{x \to 0}\left( \dfrac{\int_{0}^{x}{\dfrac{{{t}^{2}}}{\sqrt{a+2{{t}^{5}}}}dt}}{bx-e\sin x} \right).......\left( i \right)
The numerator of the LHS is a definite integral, the lower limit of which is equal to 0, and the upper limit is equal to x. Since x is tending towards zero, the upper limit and the lower limit will coincide, which means that the value of the integral and hence the numerator will approach towards zero. And the denominator will clearly tend towards zero as x tends towards zero. So the limit is of the form 00\dfrac{0}{0} which means that we can apply the L-hospital’s rule. According to the L-hospital’s rule, we need to differentiate the numerator and the denominator. Therefore, considering the numerator we have
⇒N(x)=∫0xt2a+2t5dt\Rightarrow N\left( x \right)=\int_{0}^{x}{\dfrac{{{t}^{2}}}{\sqrt{a+2{{t}^{5}}}}dt}
On differentiating both the sides, we get
⇒N′(x)=ddx(∫0xt2a+2t5dt).......(ii)\Rightarrow N'\left( x \right)=\dfrac{d}{dx}\left( \int_{0}^{x}{\dfrac{{{t}^{2}}}{\sqrt{a+2{{t}^{5}}}}dt} \right).......\left( ii \right)
Now, the derivative of a definite integral can be given as
⇒ddx(∫k1(x)k2(x)f(t)dt)=d(k2(x))dxf(k2(x))−d(k1(x))dxf(k1(x))\Rightarrow \dfrac{d}{dx}\left( \int_{{{k}_{1}}\left( x \right)}^{{{k}_{2}}\left( x \right)}{f\left( t \right)dt} \right)=\dfrac{d\left( {{k}_{2}}\left( x \right) \right)}{dx}f\left( {{k}_{2}}\left( x \right) \right)-\dfrac{d\left( {{k}_{1}}\left( x \right) \right)}{dx}f\left( {{k}_{1}}\left( x \right) \right)
Putting f(t)=t2a+2t5f\left( t \right)=\dfrac{{{t}^{2}}}{\sqrt{a+2{{t}^{5}}}}, k1(x)=0{{k}_{1}}\left( x \right)=0 and k2(x)=x{{k}_{2}}\left( x \right)=x we get
⇒ddx(∫0xt2a+2t5dt)=d(x)dx(x2a+2x5)−d(0)dx(02a+2(0)5) ⇒ddx(∫0xt2a+2t5dt)=x2a+2x5 \begin{aligned} & \Rightarrow \dfrac{d}{dx}\left( \int_{0}^{x}{\dfrac{{{t}^{2}}}{\sqrt{a+2{{t}^{5}}}}dt} \right)=\dfrac{d\left( x \right)}{dx}\left( \dfrac{{{x}^{2}}}{\sqrt{a+2{{x}^{5}}}} \right)-\dfrac{d\left( 0 \right)}{dx}\left( \dfrac{{{0}^{2}}}{\sqrt{a+2{{\left( 0 \right)}^{5}}}} \right) \\\ & \Rightarrow \dfrac{d}{dx}\left( \int_{0}^{x}{\dfrac{{{t}^{2}}}{\sqrt{a+2{{t}^{5}}}}dt} \right)=\dfrac{{{x}^{2}}}{\sqrt{a+2{{x}^{5}}}} \\\ \end{aligned}
On putting this in the equation (ii) we get
⇒N′(x)=x2a+2x5......(iii)\Rightarrow N'\left( x \right)=\dfrac{{{x}^{2}}}{\sqrt{a+2{{x}^{5}}}}......\left( iii \right)
Now, we consider the denominator from (i)
⇒D(x)=bx−esin⁡x\Rightarrow D\left( x \right)=bx-e\sin x
Differentiating both the sides, we get
⇒D′(x)=b−ecos⁡x........(iv)\Rightarrow D'\left( x \right)=b-e\cos x........\left( iv \right)
Now, using the L-Hospital’s rule, the equation (i) can be written as
⇒LHS=lim⁡x→0N′(x)D′(x)........(v)\Rightarrow LHS=\displaystyle \lim_{x \to 0}\dfrac{N'\left( x \right)}{D'\left( x \right)}........\left( v \right)
Putting (iii) and (iv) in (v) we get

& \Rightarrow LHS=\displaystyle \lim_{x \to 0}\dfrac{{{x}^{2}}}{\left( b-e\cos x \right)\sqrt{a+2{{x}^{5}}}} \\\ & \Rightarrow LHS=\dfrac{{{\left( 0 \right)}^{2}}}{\left( b-e\cos \left( 0 \right) \right)\sqrt{a+2{{\left( 0 \right)}^{5}}}} \\\ & \Rightarrow LHS=\dfrac{0}{\left( b-e \right)\sqrt{a}} \\\ \end{aligned}$$ The above limit appears to be equal to zero. But the RHS of the given equation is non-zero. This contradiction can be ended when $\left( b-e \right)$ becomes equal to zero so that the above limit again becomes the indeterminate form of $\dfrac{0}{0}$. Therefore, on equating $\left( b-e \right)$ to zero, we get $\begin{aligned} & \Rightarrow b-e=0 \\\ & \Rightarrow b=e \\\ \end{aligned}$ Putting this in (iv) we get $\begin{aligned} & \Rightarrow D'\left( x \right)=e-e\cos x \\\ & \Rightarrow D'\left( x \right)=e\left( 1-\cos x \right)......\left( vi \right) \\\ \end{aligned}$ Now, putting the equations (iii) and (vi) into (v) we get $$\Rightarrow LHS=\displaystyle \lim_{x \to 0}\dfrac{{{x}^{2}}}{e\left( 1-\cos x \right)\sqrt{a+2{{x}^{5}}}}$$ Using the algebra of limits, we can split the above limit as $$\begin{aligned} & \Rightarrow LHS=\left( \displaystyle \lim_{x \to 0}\dfrac{1}{e\sqrt{a+2{{x}^{5}}}} \right)\left( \displaystyle \lim_{x \to 0}\dfrac{{{x}^{2}}}{\left( 1-\cos x \right)} \right) \\\ & \Rightarrow LHS=\left( \dfrac{1}{e\sqrt{a+2{{\left( 0 \right)}^{5}}}} \right)\left( \displaystyle \lim_{x \to 0}\dfrac{{{x}^{2}}}{\left( 1-\cos x \right)} \right) \\\ & \Rightarrow LHS=\dfrac{1}{e\sqrt{a}}\left( \displaystyle \lim_{x \to 0}\dfrac{{{x}^{2}}}{\left( 1-\cos x \right)} \right).......\left( vii \right) \\\ \end{aligned}$$ Now, we know the trigonometric identity $$\begin{aligned} & \Rightarrow \cos 2\theta =1-2{{\sin }^{2}}\theta \\\ & \Rightarrow 1-\cos 2\theta =2{{\sin }^{2}}\theta \\\ \end{aligned}$$ Putting $$\theta =\dfrac{x}{2}$$, we get $$\begin{aligned} & \Rightarrow 1-\cos 2\left( \dfrac{x}{2} \right)=2{{\sin }^{2}}\dfrac{x}{2} \\\ & \Rightarrow 1-\cos x=2{{\sin }^{2}}\dfrac{x}{2} \\\ \end{aligned}$$ Substituting the above identity in the equation (vii) we get $$\Rightarrow LHS=\dfrac{1}{e\sqrt{a}}\left( \displaystyle \lim_{x \to 0}\dfrac{{{x}^{2}}}{2{{\sin }^{2}}\left( \dfrac{x}{2} \right)} \right)$$ Multiplying and dividing $2$ inside the bracket, we get $$\begin{aligned} & \Rightarrow LHS=\dfrac{1}{e\sqrt{a}}\left( \displaystyle \lim_{x \to 0}\dfrac{2{{x}^{2}}}{{{2}^{2}}{{\sin }^{2}}\left( \dfrac{x}{2} \right)} \right) \\\ & \Rightarrow LHS=\dfrac{1}{e\sqrt{a}}\left( \displaystyle \lim_{x \to 0}\dfrac{2{{\left( \dfrac{x}{2} \right)}^{2}}}{{{\sin }^{2}}\left( \dfrac{x}{2} \right)} \right) \\\ \end{aligned}$$ Taking $2$ outside the limit, we get $$\begin{aligned} & \Rightarrow LHS=\dfrac{2}{e\sqrt{a}}\left( \displaystyle \lim_{x \to 0}\dfrac{{{\left( \dfrac{x}{2} \right)}^{2}}}{{{\sin }^{2}}\left( \dfrac{x}{2} \right)} \right) \\\ & \Rightarrow LHS=\dfrac{2}{e\sqrt{a}}\left[ \displaystyle \lim_{x \to 0}{{\left( \dfrac{\left( \dfrac{x}{2} \right)}{\sin \left( \dfrac{x}{2} \right)} \right)}^{2}} \right] \\\ & \Rightarrow LHS=\dfrac{2}{e\sqrt{a}}\left[ \displaystyle \lim_{x \to 0}{{\left( \dfrac{\sin \left( \dfrac{x}{2} \right)}{\left( \dfrac{x}{2} \right)} \right)}^{-2}} \right] \\\ \end{aligned}$$ Substituting $\dfrac{x}{2}=t$, we get $$\begin{aligned} & \Rightarrow LHS=\dfrac{2}{e\sqrt{a}}\left[ \displaystyle \lim_{t\to 0}{{\left( \dfrac{\sin t}{t} \right)}^{-2}} \right] \\\ & \Rightarrow LHS=\dfrac{2}{e\sqrt{a}}{{\left[ \displaystyle \lim_{t\to 0}\left( \dfrac{\sin t}{t} \right) \right]}^{-2}} \\\ \end{aligned}$$ Now, we know that $$\displaystyle \lim_{t\to 0}\left( \dfrac{\sin t}{t} \right)=1$$. Substituting this above, we get $$\begin{aligned} & \Rightarrow LHS=\dfrac{2}{e\sqrt{a}}{{\left[ 1 \right]}^{-2}} \\\ & \Rightarrow LHS=\dfrac{2}{e\sqrt{a}} \\\ \end{aligned}$$ According to the given equation, we have $$\Rightarrow RHS=\dfrac{1}{\pi }$$ On equating the LHS and the RHS, we get $\Rightarrow \dfrac{2}{e\sqrt{a}}=\dfrac{1}{\pi }$ Taking reciprocals on both the sides, we get $\Rightarrow \dfrac{e\sqrt{a}}{2}=\pi $ Multiplying by $\dfrac{2}{e}$ both the sides $$\begin{aligned} & \Rightarrow \dfrac{e\sqrt{a}}{2}\left( \dfrac{2}{e} \right)=\dfrac{2}{e}\pi \\\ & \Rightarrow \sqrt{a}=\dfrac{2}{e}\pi \\\ \end{aligned}$$ Finally, taking squares on both the sides, we get $\begin{aligned} & \Rightarrow a=\dfrac{{{2}^{2}}}{{{e}^{2}}}{{\pi }^{2}} \\\ & \Rightarrow a=\dfrac{4{{\pi }^{2}}}{{{e}^{2}}} \\\ \end{aligned}$ **Hence, the value of a is equal to $\dfrac{4{{\pi }^{2}}}{{{e}^{2}}}$.** **Note:** For solving these types of questions, the concepts of determinate and indeterminate forms of limits must be very clear to us. We must note that in the question, the value of b was not given to us. So we obtained it equal to e, so that the limit does not become zero, since the RHS is non-zero.