lim _{n rightarrow infty} n^4left[frac{1}{n^5 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) આપેલ લક્ષ: $\lim _{n \rightarrow \infty} n^4 \sum_{r=0}^n \frac{1}{\left(n^2+r^2\right)^{\frac{5}{2}}}$$= \lim _{n \rightarrow \infty} \frac{1}{n}

Vedclass · Accepted Answer

$\frac{5}{6 \sqrt{2}}$

Vedclass · Answer

(C) આપેલ લક્ષ: $\lim _{n ightarrow \infty} n^4 \sum_{r=0}^n \frac{1}{\left(n^2+r^2ight)^{\frac{5}{2}}}$$= \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=0}^n \frac{n^5}{\left(n^2+r^2ight)^{\frac{5}{2}}}$$= \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=0}^n \frac{1}{\left(1+\left(\frac{r}{n}ight)^2ight)^{\frac{5}{2}}}$$= \int_0^1 \frac{1}{\left(1+x^2ight)^{\frac{5}{2}}} dx$ધારો કે $x = 	an 	heta$,તો $dx = \sec^2 	heta d	heta$. જ્યારે $x=0, 	heta=0$ અને જ્યારે $x=1, 	heta=\frac{\pi}{4}$.$I = \int_0^{\frac{\pi}{4}} \frac{\sec^2 	heta d	heta}{(\sec^2 	heta)^{\frac{5}{2}}} = \int_0^{\frac{\pi}{4}} \frac{\sec^2 	heta}{\sec^5 	heta} d	heta = \int_0^{\frac{\pi}{4}} \cos^3 	heta d	heta$$I = \int_0^{\frac{\pi}{4}} \cos 	heta (1 - \sin^2 	heta) d	heta$ધારો કે $\sin 	heta = t$,તો $\cos 	heta d	heta = dt$. જ્યારે $	heta=0, t=0$ અને જ્યારે $	heta=\frac{\pi}{4}, t=\frac{1}{\sqrt{2}}$.$I = \int_0^{\frac{1}{\sqrt{2}}} (1 - t^2) dt = \left[ t - \frac{t^3}{3} ight]_0^{\frac{1}{\sqrt{2}}}$$I = \frac{1}{\sqrt{2}} - \frac{1}{3(2\sqrt{2})} = \frac{1}{\sqrt{2}} - \frac{1}{6\sqrt{2}} = \frac{6-1}{6\sqrt{2}} = \frac{5}{6\sqrt{2}}$

$\lim _{n \rightarrow \infty} n^4\left[\frac{1}{n^5}+\frac{1}{\left(n^2+1\right)^{\frac{5}{2}}}+\frac{1}{\left(n^2+4\right)^{\frac{5}{2}}}+\frac{1}{\left(n^2+9\right)^{\frac{5}{2}}}+\ldots+\right]=$

Similar Questions

જો $a$ અને $b$ ધન પૂર્ણાંકો હોય કે જેથી $b > a$,તો $\lim_{n \to \infty} \left[ \frac{1}{na} + \frac{1}{na + 1} + \frac{1}{na + 2} + \dots + \frac{1}{nb} \right] = $

$\lim _{n \rightarrow \infty} \frac{\pi}{2 n}\left[\sin \frac{\pi}{2 n}+\sin \frac{2 \pi}{2 n}+\sin \frac{3 \pi}{2 n}+\ldots+\sin \frac{\pi}{2}\right]=$

$\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^3}\right)^{\frac{1}{n^3}}\left(1+\frac{8}{n^3}\right)^{\frac{4}{n^3}}\left(1+\frac{27}{n^3}\right)^{\frac{9}{n^3}} \ldots \left(1+\frac{n^3}{n^3}\right)^{\frac{n^2}{n^3}}\right]=$

$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x)$ ની કિંમત શોધો.

$\lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \dots + \sqrt{n}}{n^{\frac{3}{2}}} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module