For each $x \in \mathbb{R}$,let $[x]$ be the greatest integer less than or equal to $x$. Then $\lim_{x \to 0^+} \frac{x([x] + |x|) \sin [x]}{|x|}$ is equal to
- A$-\sin 1$
- B$0$
- C$1$
- D$\sin 1$
Explore More
Similar Questions
$\lim_{x \rightarrow \frac{\pi}{2}} (\tan^{2} x (\sqrt{2 \sin^{2} x + 3 \sin x + 4} - \sqrt{\sin^{2} x + 6 \sin x + 2}))$ is equal to
DifficultJEE MAIN 2022
View Solution$\lim _{n \rightarrow \infty} \frac{2^2+4^2+6^2+\ldots+(2 n)^2}{n^3} = $
$\lim _{x \rightarrow \infty} x^3 \left\{\sqrt{x^2+\sqrt{1+x^4}}-x \sqrt{2}\right\} = $
If $\lim _{n \rightarrow \infty} x^n \log _e x=0$,then $\log _x 12=$
$\lim _{x \rightarrow -3} \left( \frac{\sin ^{-1}(x+3)}{x^2+3x} \right)$ is equal to
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo