Let f be differentiable at x = 0 and f'(0) = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that $f$ is differentiable at $x = 0$ and $f'(0) = 1$. We need to evaluate the limit: $L = \lim_{h 	o 0} \frac{f(h) - f(-2h)}{h}$. By addin

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) Given that $f$ is differentiable at $x = 0$ and $f'(0) = 1$. We need to evaluate the limit: $L = \lim_{h 	o 0} \frac{f(h) - f(-2h)}{h}$. By adding and subtracting $f(0)$ in the numerator,we get: $L = \lim_{h 	o 0} \frac{f(h) - f(0) - (f(-2h) - f(0))}{h}$. $L = \lim_{h 	o 0} \left( \frac{f(h) - f(0)}{h} - \frac{f(-2h) - f(0)}{h} ight)$. $L = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} - \lim_{h 	o 0} \frac{f(-2h) - f(0)}{h}$. Since $f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = 1$,the first term is $1$. For the second term,let $k = -2h$. As $h 	o 0$,$k 	o 0$. $\lim_{h 	o 0} \frac{f(-2h) - f(0)}{h} = \lim_{k 	o 0} \frac{f(k) - f(0)}{-k/2} = -2 \lim_{k 	o 0} \frac{f(k) - f(0)}{k} = -2 f'(0) = -2(1) = -2$. Thus,$L = 1 - (-2) = 1 + 2 = 3$.

Let $f$ be differentiable at $x = 0$ and $f'(0) = 1$. Then $\lim_{h \to 0} \frac{f(h) - f(-2h)}{h} = $

Similar Questions

Find the derivative of $\frac{x^{n}-a^{n}}{x-a}$ with respect to $x$,where $a$ is a constant.

If $f(x) = \sin \left(\cosh \left(\frac{x^2+1}{x^2+2}\right)\right)$,then $f^{\prime}(1) = $

If $\frac{d}{{dx}}\left[ {\frac{{2{x^3} + 3{x^2} + x - 3}}{{{x^2} + x - 2}}} \right] = A + \frac{B}{{{{(x - 1)}^2}}} + \frac{C}{{{{(x + 2)}^2}}}$ then $(A - B + C)$ is

If $f(x)=|x-1|+|x-2|$,then $f^{\prime}(-2023)+f^{\prime}\left(\frac{2024}{2023}\right)+f^{\prime}(2023)=$

If $f(x) = 1 + x + x^2 + \ldots + x^{1000}$,then $f^{\prime}(-1) = $ . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module