Question

The value of \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right) is –
(a) $0$
(b) $1$
(c) $\dfrac{1}{2}$
(d) Non existent

Answer 1

Hint: The given expression is a combination of integration, differentiation and limit. So, first solve the innermost function and then proceed outwards. Here first solve the integration part, then apply the derivative and at last apply the limit.

Complete step-by-step answer:
Given,
\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)
First we will solve the integration part.
\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{{{r}^{2}}-1}dr
\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{r\left( {{r}^{2}} \right)}{{{r}^{2}}-1}dr\ldots \left( i \right)
Let,
$u={{r}^{2}}-1\Rightarrow {{r}^{2}}=u+1$
$\Rightarrow \dfrac{du}{dr}=2r$
$\Rightarrow dr=\dfrac{1}{2r}du$
Substituting these values in equation (i), we get
\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{r\left( u+1 \right)}{u}\left( \dfrac{1}{2r}du \right)
\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{\left( u+1 \right)}{u}\left( \dfrac{1}{2}du \right)
\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{\left( u+1 \right)}{u}du
On expanding, we get
\Rightarrow \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\left( \dfrac{u}{u}+\dfrac{1}{u} \right)du
\Rightarrow \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\left( 1+\dfrac{1}{u} \right)du
This can be also written as,
\Rightarrow \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\left[ \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,1du+\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{1}{u}du \right]
But we know, $\mathop{\int }^{}\dfrac{1}{u}du=\ln u$ , so we get
\Rightarrow \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\left[ u+\ln u \right]_{\sqrt{3}}^{\sqrt{x}}
Substituting back the values of ‘u’ and ‘du’, we get
\Rightarrow \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\left[ \left( {{r}^{2}}-1 \right)+\ln \left( {{r}^{2}}-1 \right) \right]_{\sqrt{3}}^{\sqrt{x}}
Applying the values, we get
\Rightarrow \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\left[ \left( \left( {{\left( \sqrt{x} \right)}^{2}}-1 \right)+\ln \left( {{\left( \sqrt{x} \right)}^{2}}-1 \right) \right)-\left( \left( {{\left( \sqrt{3} \right)}^{2}}-1 \right)+\ln \left( {{\left( \sqrt{3} \right)}^{2}}-1 \right) \right) \right]
\Rightarrow \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\left[ \left( \left( x-1 \right)+\ln \left( x-1 \right) \right)-\left( \left( 3-1 \right)+\ln \left( 3-1 \right) \right) \right]
\Rightarrow \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr=\dfrac{1}{2}\left[ \left( \left( x-1 \right)+\ln \left( x-1 \right) \right)-\left( 2+\ln 2 \right) \right]
Now we will apply the derivative, we get
\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{d}{dx}\left\\{ \dfrac{1}{2}\left[ \left( \left( x-1 \right)+\ln \left( x-1 \right) \right)-\left( 2+\ln 2 \right) \right] \right\\}
\Rightarrow \dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\\{ \dfrac{d}{dx}\left[ \left( x-1 \right)+\ln \left( x-1 \right) \right]-\dfrac{d}{dx}\left[ 2+\ln 2 \right] \right\\}
But differentiation of constant is a constant, so
\Rightarrow \dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\\{ \dfrac{d}{dx}\left[ \left( x-1 \right)+\ln \left( x-1 \right) \right]-0 \right\\}
This can be written as,
\Rightarrow \dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\\{ \dfrac{d}{dx}\left[ x \right]-\dfrac{d}{dx}\left[ 1 \right]+\dfrac{d}{dx}\left[ \ln \left( x-1 \right) \right] \right\\}
Solving we get,
\Rightarrow \dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\\{ 1-0+\dfrac{d}{dx}\left[ \ln \left( x-1 \right) \right] \right\\}
Differentiation of $\ln y=\dfrac{1}{y}.\dfrac{dy}{dx}$ , so we get
\Rightarrow \dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\\{ 1+\dfrac{1}{x-1}.\dfrac{d}{dx}\left[ x-1 \right] \right\\}
\Rightarrow \dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\\{ 1+\dfrac{1}{x-1}.\left( 1 \right) \right\\}
\Rightarrow \dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\\{ \dfrac{\left( x-1 \right)+1}{x-1} \right\\}
\Rightarrow \dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\\{ \dfrac{x}{x-1} \right\\}
Now we will apply limit, we get
\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\mathop{\overset{\sqrt{x}}{\mathop{\int }}\,}}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{2}\left\\{ \dfrac{x}{x-1} \right\\}
\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{2}\left\\{ \dfrac{1}{\dfrac{\left( x-1 \right)}{x}} \right\\}
\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{2}\left\\{ \dfrac{1}{1-\dfrac{1}{x}} \right\\}
Now we will apply the limit, we get
\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\\{ \dfrac{1}{1-\dfrac{1}{\infty }} \right\\}
We know, $\dfrac{1}{\infty }\approx 0$ ,
\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}\left\\{ \dfrac{1}{1-0} \right\\}
\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{d}{dx}\left( \underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr \right)=\dfrac{1}{2}
Hence, the correct option for the given question is option (c).
Answer is Option (c)

Note: In the solution the following equation,
\underset{\sqrt{3}}{\overset{\sqrt{x}}{\mathop \int }}\,\dfrac{{{r}^{3}}}{\left( r+1 \right)\left( r-1 \right)}dr
It can be solved by applying a partial fraction decomposition method too. The answer won’t differ.