$\lim _{x \rightarrow-\infty} \frac{3|x|^3-x^2+2|x|-5}{-5|x|^3+3 x^2-2|x|+7} = $

  • A
    $\frac{3}{5}$
  • B
    $\frac{-5}{7}$
  • C
    $\frac{5}{7}$
  • D
    $\frac{-3}{5}$

Explore More

Similar Questions

$\lim _{x \rightarrow 0} \frac{\cos 4 x-4 \cos 2 x+3}{x^4} = $

मान लीजिए $[x]$ उस सबसे बड़े पूर्णांक को दर्शाता है जो $x$ से अधिक नहीं है। यदि $l_1 = \lim_{x \rightarrow 2^{+}} (x^2 + [x])$,$l_2 = \lim_{x \rightarrow 3^{-}} (2x - [x])$ और $l_3 = \lim_{x \rightarrow \frac{\pi}{2}} \left( \frac{\cos x}{x - \frac{\pi}{2}} \right)$ है,तो:

$\lim _{x \rightarrow \infty} x^3 \left\{\sqrt{x^2+\sqrt{1+x^4}}-x \sqrt{2}\right\} = $

$\mathop {\lim }\limits_{x \to 0} \left[ \frac{\sqrt{a + x} - \sqrt{a - x}}{x} \right]$ का मान है

दिए गए सीमा (limit) का मूल्यांकन करें: $\mathop {\lim }\limits_{x \to 0} \frac{\cos x}{\pi - x}$

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