$\lim _{x \rightarrow 0} \left( \frac{1}{x} \ln \sqrt{\frac{1+x}{1-x}} \right)$ का मान है

  • A
    $1/2$
  • B
    $0$
  • C
    $1$
  • D
    अस्तित्व में नहीं है

Explore More

Similar Questions

मान लीजिए कि $S$ सभी $(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}$ का समुच्चय है ताकि $\lim _{x \rightarrow \infty} \frac{\sin(x^2)(\log_e x)^\alpha \sin(1/x^2)}{x^{\alpha \beta}(\log_e(1+x))^\beta} = 0$ हो। तो निम्नलिखित में से कौन सा (से) सही है (हैं)?

$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 8x + 3} - \sqrt {{x^2} + 4x + 3} ) = $

$\mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt \pi - \sqrt {{{\cos }^{ - 1}}x} }}{{\sqrt {x + 1} }}$ का मान ज्ञात कीजिए।

$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n k^2 x = $

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

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