Question

Evaluate the limit \[\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{n}{{{n}^{2}}+{{1}^{2}}}+\dfrac{n}{{{n}^{2}}+{{1}^{2}}}+\dfrac{n}{{{n}^{2}}+{{1}^{2}}}+\,.....\,+\,\dfrac{1}{5n} \right)\].
(a) $\dfrac{\pi }{4}$
(b) ${{\tan }^{-1}}2$
(c) ${{\tan }^{-1}}3$
(d) $\dfrac{\pi }{2}$

Answer 1

Hint: To solve this problem we need to apply reverse of limit as sum because here we cannot apply direct limit to given function as how many number of terms are there in between, is not defined. Reverse of limit is the method through which we will convert the given limit to the integral form then solve the integral to find the answer.

Complete step by step answer:
We have to evaluate the limit,
\[\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{n}{{{n}^{2}}+{{1}^{2}}}+\dfrac{n}{{{n}^{2}}+{{1}^{2}}}+\dfrac{n}{{{n}^{2}}+{{1}^{2}}}+\,.....\,+\,\dfrac{1}{5n} \right)\]
We can also express the given limit as,
$\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{r=2n}{\left( \dfrac{n}{{{n}^{2}}+{{r}^{2}}} \right)}$
\[=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{r=2n}{\dfrac{1}{n}\left( \dfrac{1}{1+{{\left( \dfrac{r}{n} \right)}^{2}}} \right)}\]
Now, we have to convert above expression to integral form (reverse of limit as sum) by changing :
$\begin{align}
  & \dfrac{1}{n}\to dx \\
 & and, \\
 & \dfrac{r}{n}\to x \\
\end{align}$
Lower limit = $\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{r}{n} \right)$= 0 , as n→∞ and r is a finite number and we know anything \[\dfrac{finite}{infinite}\]= 0.
Upper limit = 2 i.e. when r = 2n in $\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{r}{n} \right)$ we get limit as 2.
Hence, integral form of given expression we get as,
\[I=\int\limits_{0}^{2}{\dfrac{1}{1+{{x}^{2}}}dx}\]
And we know that $\int{\dfrac{1}{1+{{x}^{2}}}dx={{\tan }^{-1}}x+c}$, so we get
$I=\left[ {{\tan }^{-1}}x \right]_{0}^{2}$
$\begin{align}
  & I={{\tan }^{-1}}2-{{\tan }^{-1}}0 \\
 & I={{\tan }^{-1}}2 \\
\end{align}$
Hence our answer matches with option (b), so option (b) is the correct answer.

Note:
To solve this problem you will have to learn the reverse of the limit as a sum properly in order to understand what variables need to be changed in a given equation. And,
You should remember Integration of $\int{\dfrac{1}{1+{{x}^{2}}}dx={{\tan }^{-1}}x+c}$ for future references. And also when you have no idea of the number of terms that are inside the limit then usually that kind of problems are solved by reverse of the limit method.