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The problem is as follows: $f$ is differentiable at $x$. Show that $\lim_{h\rightarrow 0}\frac{f^2(x+3h)-f^2(x-h)}{h}$ exists and find its value. Note that $f^2(a)$ just means $[f(a)]^2$.

Well, basis calculus tells me the answer should be $2f'(x)f(x)$.

Here's my attempt: using the fact that $a^2-b^2=(a-b)(a+b)$, I get that the fraction turns into $\frac{f(x+3h)-f(x-h)}{h}\cdot (f(x+3h)+f(x-h))$. I know since $f$ is differentiable at $x$, it's continuous in a neighborhood around $x$, and so the right product just turns into $2f(x)$ as $h\rightarrow 0$. The left product is basically $f'(x)$, but I'm worried abound $f$ being evaluated at $x+3h$ and $x-h$, rather than strictly $x+h$ and $x$. How do I deal with this? Or can I just say that the left product is $f'(x)$ has $h\rightarrow 0$? Thanks.

NoChance
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user0424
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    Note that $h\to 0$ implies $3h \to 0$ – tomate Apr 25 '23 at 02:01
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    Remember intuitively, $f'(x)$ denotes the approximate slope of secants near $x$. The equation for slope is $\frac{\Delta y}{\Delta x}$, and $\Delta x = (x+3h) - (x-h) = 4h \neq h$. (Note that as an alternative, you can also apply L'Hopital's rule here with respect to $h$, with $x$ held constant) – Brian Moehring Apr 25 '23 at 02:01

1 Answers1

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Let $g = f^2$.

We are looking for

$$L = \lim_{h \to 0} \frac{g (x + 3 h) - g (x - h)}{h}$$

Let $h' = 4 h$; note that $h \to 0 \implies h' \to 0$. Thus,

$$L = 4 \lim_{h' \to 0} \frac{g \left( \left( x - \frac{1}{4} h' \right) + h' \right) - g \left( x - \frac{1}{4} h' \right)}{h'}$$

But $\lim_{h' \to 0} x - \frac{1}{4} h' = x$, so

$$L = 4 \lim_{h' \to 0} \frac{g (x + h') - g (x)}{h'} = 4 g' (x)$$

By basic rules of calculus, we know that $g' = 2 f f'$, so

$$L = \boxed{8 f (x) f' (x)}$$

K. Jiang
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