$\mathop {\lim }\limits_{x \to 0} \frac{{{4^x} - {9^x}}}{{x({4^x} + {9^x})}} = $

  • A
    $\log \left( {\frac{2}{3}} \right)$
  • B
    $\frac{1}{2}\log \left( {\frac{3}{2}} \right)$
  • C
    $\frac{1}{2}\log \left( {\frac{2}{3}} \right)$
  • D
    $\log \left( {\frac{3}{2}} \right)$

Explore More

Similar Questions

Evaluate the limit: $\lim _{x \rightarrow 0} \frac{e^x-e^{\sin x}}{2(x-\sin x)}$

Evaluate the limit: $\lim _{x \rightarrow 1} \left[\frac{\sqrt{x}-1}{\log x}\right]$

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\sin x}} - 1}}{x} = $

If $f(0) = 2,$ then $\mathop {\lim }\limits_{x \to 0} \frac{{\int\limits_0^x {\left( {tf(x) + xf(t)} \right)dt} }}{{{x^2}}}$ is equal to -

Let $\alpha$ be a positive real number. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: (\alpha, \infty) \rightarrow \mathbb{R}$ be the functions defined by $f(x) = \sin \left(\frac{\pi x}{12}\right)$ and $g(x) = \frac{2 \log_{e}(\sqrt{x}-\sqrt{\alpha})}{\log_{e}(e^{\sqrt{x}}-e^{\sqrt{\alpha}})}$. Then the value of $\lim_{x \rightarrow \alpha^{+}} f(g(x))$ is

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