mathop {lim }limits_{x to 0} x^2(1+2+3+...+[f | vedclass

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

Question

(B) Let $t = \frac{1}{|x|}$. As $x 	o 0$,$t 	o \infty$.The expression becomes $\mathop {\lim }\limits_{t 	o \infty } \frac{1}{t^2} (1 + 2 + 3 + ...

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) Let $t = \frac{1}{|x|}$. As $x 	o 0$,$t 	o \infty$.The expression becomes $\mathop {\lim }\limits_{t 	o \infty } \frac{1}{t^2} (1 + 2 + 3 + ... + [t])$.Using the sum formula for the first $n$ natural numbers,$1+2+...+n = \frac{n(n+1)}{2}$,we have:$\mathop {\lim }\limits_{t 	o \infty } \frac{[t]([t]+1)}{2t^2}$.Since $\mathop {\lim }\limits_{t 	o \infty } \frac{[t]}{t} = 1$,the limit is $\mathop {\lim }\limits_{t 	o \infty } \frac{1}{2} \cdot \frac{[t]}{t} \cdot \frac{[t]+1}{t} = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 0} x^2(1+2+3+...+[\frac{1}{|x|}])$ is equal to (where $[.]$ denotes the greatest integer function).

Similar Questions

$\lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4} = $

$\mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt \pi - \sqrt {{{\cos }^{ - 1}}x} }}{{\sqrt {x + 1} }}$ is equal to

$\mathop {\lim }\limits_{x \to 3} [x] = $,(where $[.]$ denotes the greatest integer function)

$\lim _{x \rightarrow 0} \frac{\sqrt{x^2+100}-10}{x^2} = $

If $x$ is a real number in $[0, 1]$,then the value of $\lim_{m \to \infty} \lim_{n \to \infty} [1 + \cos^{2m}(n! \pi x)]$ is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module