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

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(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

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$3$

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(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} = $

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