The limit L=limn→∞⁡∑r=4n-4n/(n^2+r^2) satisfies the relation

Correct answer: (C) e^(tan⁡L)<2e. Given limits is L=limn→∞⁡∑r=4n-41/(1+r^n2)⋅1/n =∫01dx/(1+x^2)=tan^(-1)⁡x01=(π)/4 Hence, e^(tanπ4)=e<2e

Q49 / 116 · NTA Abhyas (JEE Main) 2020 · Mathematics · Definite Integration

The limit $L = \lim_{n \to \infty} \sum_{r = 4}^{n - 4} \frac{n}{n^{2} + r^{2}}$ satisfies the relation

A
$e^{L} > e$
B
$e^{L} > 3$
Correct answer
C
$e^{\tan L} < 2 e$
D
$\frac{π}{L} < 1$

Solution

Given limits is $L = \lim_{n \to \infty} \sum_{r = 4}^{n - 4} \frac{1}{1 + {(\frac{r}{n})}^{2}} \cdot \frac{1}{n}$
$= \int_{0}^{1} \frac{d x}{1 + x^{2}} = {[\tan^{- 1} x]}_{0}^{1} = \frac{π}{4}$

Hence, $e^{\tan \frac{π}{4}} = e < 2 e$

Hence, the correct answer is option: 'C'.

The limit L=limn→∞⁡∑r=4n-4nn2+r2 satisfies the relation

Solution

More Definite Integration Questions

The limit $L = \lim_{n \to \infty} \sum_{r = 4}^{n - 4} \frac{n}{n^{2} + r^{2}}$ satisfies the relation