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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