Question

$\underset{n\to \infty }{lim}{\sum }_{r=1}^n\frac{n}{n^2+r^2{x}^2},x>0$is equal to :

• A. $ta{n}^{-1}x$
• B. $xta{n}^{-1}x$
• C. $\frac{ta{n}^{-1}x}{x}$
• D. $\frac{ta{n}^{-1}x}{x^2}$

Answer 1

The correct option is C $\frac{ta{n}^{-1}x}{x}$

$\underset{n\to \infty }{lim}{\sum }_{r=1}^n\frac{n}{n^2+r^2{x}^2}=\underset{n\to \infty }{lim}{\sum }_{r=1}^n\frac{1}{n(1+(\frac{r}{n}{)}^2{x}^2)}={\int }_0^1\frac{dt}{1+t^2{x}^2}=\frac{1}{x^2}{\int }_0^1\frac{dt}{{\left(\frac{1}{x}\right)}^2+t^2}=\frac{1}{x^2}.\frac{1}{1/x}{\left[ta{n}^{-1}\left(\frac{1}{1/x}\right)\right]}_0^1=\frac{1}{x}ta{n}^{-1}x=\frac{ta{n}^{-1}x}{x}$