Lim _ n , , _ k = 1 ^n n n^2 + k^2 x^2 , x > 0 is equal to - Vidyadip

MCQ

$\mathop {Lim}\limits_{n \to \infty } \,\,\sum\limits_{k = 1}^n {\frac{n}{{{n^2} + {k^2}{x^2}}}} $,$ x > 0$ is equal to

A
$x tan^{-1}(x)$
B
$tan^{-1}(x)$
✓
$\frac{{{{\tan }^{ - 1}}(x)}}{x}$
D
$\frac{{{{\tan }^{ - 1}}(x)}}{{{x^2}}}$

✓

Answer

Correct option: C.

$\frac{{{{\tan }^{ - 1}}(x)}}{x}$

c
$T_{n}=\frac{n}{n^{2}+k^{2} x^{2}}=\frac{1}{n\left[1+(k / n)^{2} x^{2}\right]}$

$S=\frac{1}{n} \sum_{k=1}^{n} \frac{1}{1+\underbrace{(k / n)^{2} x^{2}}}_{1}=\int_{0}^{1} \frac{d t}{1+t^{2} x^{2}}$

$=\frac{1}{x^{2}} \int_{0}^{1} \frac{d t}{t^{2}+\left(1 / x^{2}\right)}=\left[\frac{1}{x} \tan ^{-1}(t x)\right]_{0}^{1}=\frac{\tan ^{-1}(x)}{x}$

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