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

Evaluate the limit: $\mathop {\lim }\limits_{x \to \infty } [x({a^{1/x}} - 1)]$,where $a > 1$.

$\lim _{y \rightarrow 1}\left(\frac{1}{y^2-1}-\frac{2}{y^4-1}\right)=$

$\mathop {\lim }\limits_{x \to 0} \left( \frac{x}{\tan^{-1} 2x} \right) = $

$\lim _{x \rightarrow 0} \frac{(3^{2x}-\sqrt{x+1}) \sin 5x}{1-\cos 4x} =$

If $\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{{a^{1/x}} + b}}{c}} \right)^x} = d$ (where $d$ is a non-zero finite value),then $(b + 1) \log_a d$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module