Let $f(x)$ be differentiable at $x = h$. Then $\lim_{x \to h} \frac{(x + h)f(x) - 2hf(h)}{x - h}$ is equal to

  • A
    $f(h) + 2hf'(h)$
  • B
    $2f(h) + hf'(h)$
  • C
    $hf(h) + 2f'(h)$
  • D
    $hf(h) - 2f'(h)$

Explore More

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{x{e^x} - \log (1 + x)}}{{{x^2}}}$ equals

Let $a > 0$ be a real number. Then the limit $\lim _{x \rightarrow 2} \frac{a^x+a^{3-x}-\left(a^2+a\right)}{a^{3-x}-a^{x / 2}}$ is

$\mathop {\lim }\limits_{x \to 1} {\left( {\frac{4}{\pi }{{\tan }^{ - 1}}x} \right)^{\frac{1}{{({x^2} - 1)}}}}$ is equal to -

$\mathop {\lim }\limits_{x \to a} \frac{{\cos x - \cos a}}{{\cot x - \cot a}} = $

If $\lim _{x \rightarrow 0} \frac{a x e^{x}-b \log (1+x)}{x^{2}}=3$, then the values of $a$ and $b$ are, respectively:

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