mathop {Lim}limits_{n to infty } ,,sumlimits_ | vedclass

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(C) We are given the limit $L = \mathop {Lim}\limits_{n 	o \infty } \sum\limits_{k = 1}^n \frac{n}{n^2 + k^2 x^2}$.Divide the numerator and denominat

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We are given the limit $L = \mathop {Lim}\limits_{n 	o \infty } \sum\limits_{k = 1}^n \frac{n}{n^2 + k^2 x^2}$.Divide the numerator and denominator by $n^2$:$L = \mathop {Lim}\limits_{n 	o \infty } \frac{1}{n} \sum\limits_{k = 1}^n \frac{1}{1 + (k/n)^2 x^2}$.This is a Riemann sum of the form $\int_0^1 f(t) dt$ where $t = k/n$ and $dt = 1/n$ as $n 	o \infty$.Thus,$L = \int_0^1 \frac{1}{1 + t^2 x^2} dt$.Let $u = tx$,then $du = x dt$,so $dt = du/x$. When $t=0, u=0$ and when $t=1, u=x$.$L = \int_0^x \frac{1}{1 + u^2} \cdot \frac{du}{x} = \frac{1}{x} [	an^{-1}(u)]_0^x$.$L = \frac{1}{x} (	an^{-1}(x) - 	an^{-1}(0)) = \frac{	an^{-1}(x)}{x}$.

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

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