Question

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

• A. 3
• B. 2
• C. 1
• D. -1

Answer 1

Answer: (A)

3

Answer 2

To evaluate the limit $\lim_{h \to 0} \frac{f(h) - f(-2h)}{h}$, we can use the definition of the derivative.

Since $f$ is differentiable at $x=0$, we can rewrite the expression by adding and subtracting $f(0)$ in the numerator:

$\lim_{h \to 0} \frac{f(h) - f(-2h)}{h} = \lim_{h \to 0} \frac{f(h) - f(0) - (f(-2h) - f(0))}{h}$

This can be split into two separate limits:

$\lim_{h \to 0} \frac{f(h) - f(0)}{h} - \lim_{h \to 0} \frac{f(-2h) - f(0)}{h}$

The first limit is the definition of the derivative of $f$ at $x=0$:

$\lim_{h \to 0} \frac{f(h) - f(0)}{h} = f'(0) = 1$

For the second limit, let $k = -2h$. As $h \to 0$, $k \to 0$.

$\lim_{h \to 0} \frac{f(-2h) - f(0)}{h} = \lim_{k \to 0} \frac{f(k) - f(0)}{-k/2} = \lim_{k \to 0} -2 \frac{f(k) - f(0)}{k} = -2f'(0) = -2$

Substituting these back into the expression:

$f'(0) - (-2f'(0)) = f'(0) + 2f'(0) = 3f'(0) = 3(1) = 3$

Therefore, the limit is 3.