Question

Find the value of the given limit $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{r=1}^{n}{\sqrt{\dfrac{n+r}{n-r}}}$?
(a) $\dfrac{\pi }{2}$,
(b) $2\pi$,
(c) $\dfrac{\pi }{2}-1$,
(d) $\dfrac{\pi }{2}+1$.

Answer 1

We start solving the problem by assigning a variable for the value of the given limit. We then rearrange the limit which resembles to use the integral as a limit of summation rule $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{r=1}^{n}{f\left( \dfrac{r}{n} \right)}=\int\limits_{0}^{1}{f\left( x \right)dx}$. We then make arrangements inside the square root and find the integrals. We then substitute the limits and make necessary calculations to get the required value of the limit.

Complete step-by-step solution:
According to the problem, we need to find the value of the given limit $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{r=1}^{n}{\sqrt{\dfrac{n+r}{n-r}}}$. Let us assume the value be ‘z’.
Now, we have $z=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{r=1}^{n}{\sqrt{\dfrac{n+r}{n-r}}}$.
$\Rightarrow z=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{r=1}^{n}{\sqrt{\dfrac{n\left( 1+\dfrac{r}{n} \right)}{n\left( 1-\dfrac{r}{n} \right)}}}$.
$\Rightarrow z=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{r=1}^{n}{\sqrt{\dfrac{\left( 1+\dfrac{r}{n} \right)}{\left( 1-\dfrac{r}{n} \right)}}}$ -------(1).
We can see that the given resembles the form which satisfies the integral as a limit of summation rule.
i.e., $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{r=1}^{n}{f\left( \dfrac{r}{n} \right)}=\int\limits_{0}^{1}{f\left( x \right)dx}$. Comparing this with equation (1), we get $f\left( x \right)=\sqrt{\dfrac{1+x}{1-x}}$.
Using the integral as a limit of summation, we get the value of z as $z=\int\limits_{0}^{1}{\sqrt{\dfrac{1+x}{1-x}}dx}$.
Let us multiply the numerator and denominator with
$\Rightarrow z=\int\limits_{0}^{1}{\sqrt{\dfrac{1+x}{1-x}\times \dfrac{1+x}{1+x}}dx}$.
$\Rightarrow z=\int\limits_{0}^{1}{\sqrt{\dfrac{{{\left( 1+x \right)}^{2}}}{\left( 1-x \right)\times \left( 1+x \right)}}dx}$.
We know that $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$.
$\Rightarrow z=\int\limits_{0}^{1}{\dfrac{\sqrt{{{\left( 1+x \right)}^{2}}}}{\sqrt{\left( 1-{{x}^{2}} \right)}}dx}$.
$\Rightarrow z=\int\limits_{0}^{1}{\dfrac{1+x}{\sqrt{1-{{x}^{2}}}}dx}$.
We know that $\int{\left( f\left( x \right)+g\left( x \right) \right)dx=\int{f\left( x \right)dx}+}\int{g\left( x \right)dx}$.
$\Rightarrow z=\int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}+\int\limits_{0}^{1}{\dfrac{x}{\sqrt{1-{{x}^{2}}}}dx}$.
Let us assume the integral ${{I}_{1}}=\int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}$ and ${{I}_{2}}=\int\limits_{0}^{1}{\dfrac{x}{\sqrt{1-{{x}^{2}}}}dx}$.
$\Rightarrow z={{I}_{1}}+{{I}_{2}}$ --------(2).
Let us find the values of ${{I}_{1}}$ and ${{I}_{2}}$ first and then add them together to get the required value of the limit.
We have ${{I}_{1}}=\int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}$ -----(3).
Let us assume $x=\sin t$ --------(4). Let us differentiate on both sides.
$\Rightarrow d\left( x \right)=d\left( \sin x \right)$.
$\Rightarrow dx=\cos tdt$ --------(5).
Let us find the upper and lower limits for the integral in equation (3).
So, we have upper limit as $x=1\Rightarrow \sin t=1$.
$\Rightarrow t={{\sin }^{-1}}\left( 1 \right)$.
$\Rightarrow t=\dfrac{\pi }{2}$ is upper limit -------(6).
So, we have lower limit as $x=0\Rightarrow \sin t=0$.
$\Rightarrow t={{\sin }^{-1}}\left( 0 \right)$.
$\Rightarrow t=0$ is lower limit --------(7).
We substitute equations (4), (5), (6), and (7) in equation (3).
$\Rightarrow {{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{1}{\sqrt{1-{{\left( \sin t \right)}^{2}}}}\left( \cos tdt \right)}$.
$\Rightarrow {{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos t}{\sqrt{1-{{\sin }^{2}}t}}dt}$.
We know that $1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta$.
$\Rightarrow {{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos t}{\sqrt{{{\cos }^{2}}t}}dt}$.
$\Rightarrow {{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos t}{\cos t}dt}$.
$\Rightarrow {{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{1dt}$.
We know that $\int{adx=ax+C}$ and $\int\limits_{a}^{b}{{{f}^{'}}\left( x \right)dx=\left[ f\left( x \right) \right]_{a}^{b}}$.
$\Rightarrow {{I}_{1}}=\left[ t \right]_{0}^{\dfrac{\pi }{2}}$.
$\Rightarrow {{I}_{1}}=\left( \dfrac{\pi }{2}-0 \right)$.
$\Rightarrow {{I}_{1}}=\dfrac{\pi }{2}$ --------(8).
Now, we have ${{I}_{2}}=\int\limits_{0}^{1}{\dfrac{x}{\sqrt{1-{{x}^{2}}}}dx}$ -------(9).
Let us assume $x=\sin t$ ---------(10). Let us differentiate on both sides.
$\Rightarrow d\left( x \right)=d\left( \sin x \right)$.
$\Rightarrow dx=\cos tdt$ -------(11).
Let us find the upper and lower limits for the integral in equation (9).
So, we have upper limit as $x=1\Rightarrow \sin t=1$.
$\Rightarrow t={{\sin }^{-1}}\left( 1 \right)$.
$\Rightarrow t=\dfrac{\pi }{2}$ is upper limit -------(12).
So, we have lower limit as $x=0\Rightarrow \sin t=0$.
$\Rightarrow t={{\sin }^{-1}}\left( 0 \right)$.
$\Rightarrow t=0$ is lower limit -------(13).
We substitute equations (10), (11), (12) and (13) in equation (9).
$\Rightarrow {{I}_{2}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin t}{\sqrt{1-{{\left( \sin t \right)}^{2}}}}\left( \cos tdt \right)}$.
$\Rightarrow {{I}_{2}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin t\times \cos t}{\sqrt{1-{{\sin }^{2}}t}}dt}$.
We know that $1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta$.
$\Rightarrow {{I}_{2}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin t\times \cos t}{\sqrt{{{\cos }^{2}}t}}dt}$.
$\Rightarrow {{I}_{2}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin t\times \cos t}{\cos t}dt}$.
$\Rightarrow {{I}_{2}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\sin tdt}$.
We know that $\int{\sin xdx=-\cos x+C}$ and $\int\limits_{a}^{b}{{{f}^{'}}\left( x \right)dx=\left[ f\left( x \right) \right]_{a}^{b}}$.
$\Rightarrow {{I}_{2}}=\left[ -\cos t \right]_{0}^{\dfrac{\pi }{2}}$.
$\Rightarrow {{I}_{2}}=\left( -\cos \left( \dfrac{\pi }{2} \right)-\left( -\cos \left( 0 \right) \right) \right)$.
$\Rightarrow {{I}_{2}}=\left( -0-\left( -1 \right) \right)$.
$\Rightarrow {{I}_{2}}=1$ --------(14).
Let us substitute equations (8) and (14) in equation (2).
So, we get $z=\dfrac{\pi }{2}+1$.
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{r=1}^{n}{\sqrt{\dfrac{n+r}{n-r}}}=\dfrac{\pi }{2}+1$.
So, we have found the value of the limit $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{r=1}^{n}{\sqrt{\dfrac{n+r}{n-r}}}$ as $\dfrac{\pi }{2}+1$.
The correct option for the given problem is (c).

Note: We should not confuse with the formulas while performing the integrations. We cannot get these problems directly by applying the given limit as the summation tends to infinity when we apply the limit. We should not make calculation mistakes while solving this problem. Whenever we get this type of problem, we first try to convert them to the form that we can apply integration as a limit of the summation rule properly.