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