$\mathop {\lim }\limits_{x \to 0} \left( \frac{\sin x - x + \frac{x^3}{6}}{x^5} \right) = $

  • A
    $1/120$
  • B
    $-1/120$
  • C
    $1/20$
  • D
    None of these

Explore More

Similar Questions

Evaluate $\mathop {\lim }\limits_{x \to \infty } \frac{{\log {x^n} - [x]}}{{[x]}},$ where $n \in N$ and $[x]$ denotes the greatest integer less than or equal to $x$.

If $\mathop {\lim }\limits_{x \to a} \frac{{{a^x} - {x^a}}}{{{x^x} - {a^a}}} = - 1$,then

$\lim _{x \rightarrow 0} \frac{x^2 2^x-x^2 \sin x-x^2}{3^x+\cos x-3^x \cos x-1}=$

$\mathop {Limit}\limits_{x \to 0^+} \frac{1}{x\sqrt{x}} \left( a \tan^{-1} \frac{\sqrt{x}}{a} - b \tan^{-1} \frac{\sqrt{x}}{b} \right)$ has the value equal to

Evaluate the limit: $\lim _{x \rightarrow \pi / 6} \frac{3 \sin x-\sqrt{3} \cos x}{6 x-\pi}$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo