mathop {lim }limits_{n to infty } frac{1}{n}s | vedclass

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(B) Let $L = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n}\sum\limits_{r = 1}^{2n} {\frac{r}{{\sqrt {{n^2} + {r^2}} }}} $.We can rewrite the expr

Vedclass · Accepted Answer

$-1 + \sqrt{5}$

Vedclass · Answer

(B) Let $L = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n}\sum\limits_{r = 1}^{2n} {\frac{r}{{\sqrt {{n^2} + {r^2}} }}} $.We can rewrite the expression as $L = \mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^{2n} {\frac{1}{n} \cdot \frac{r/n}{{\sqrt {1 + {{(r/n)}^2}} }}} $.Using the definition of a definite integral as the limit of a sum,$\mathop {\lim }\limits_{n 	o \infty } \frac{1}{n}\sum\limits_{r = 1}^{kn} {f(r/n)} = \int_0^k {f(x)dx}$.Here,$f(x) = \frac{x}{{\sqrt {1 + {x^2}} }}$ and the upper limit is $k = 2$.So,$L = \int_0^2 {\frac{x}{{\sqrt {1 + {x^2}} }}} dx$.Let $u = 1 + x^2$,then $du = 2x dx$,or $x dx = \frac{1}{2} du$.When $x = 0$,$u = 1$. When $x = 2$,$u = 5$.$L = \int_1^5 {\frac{1}{{2\sqrt u }}} du = \left[ {\sqrt u } ight]_1^5 = \sqrt{5} - 1$.

$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^{2n} {\frac{r}{{\sqrt {{n^2} + {r^2}} }}} $ equals

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