Right, so this was a question I randomly pondered up while I was watching through a BlackPenRedPen video, in which he shows off a question similar to the one I'm about to ask, except with f(x) = $x^2$ instead. Right, here it is:
Let f(x) = $x^3$.
a) Prove that $f'(x) = 3x^2$, using the definition of a derivative.
b) Prove part (a), through an epsilon delta definition.
Part (a) is easy enough. I just went as follows:
$$\lim_{h\to 0} \frac{f(x+h) – f(x)}{h}$$
$$ = \lim_{h\to 0} \frac{(x+h)^3 – x^3}{h} $$
$$ = \lim_{h\to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3-x^3}{h} $$
$$ = \lim_{h\to 0} \frac{3x^2h + 3xh^2 + h^3}{h} $$
$$ = \lim_{h\to 0} \frac{h(3x^2 + 3xh + h^2)}{h} $$
$$ = \lim_{h\to 0} (3x^2 + 3xh + h^2) $$
$$ = \lim_{h\to 0} 3x^2 + h(3x + h) $$
$$ = 3x^2 $$
However, my problem begins with part (b), A.K.A the epsilon delta proof. Here's how I started:
Proof: Let $\epsilon$ > 0. Set $\delta$ = (BLANK). Suppose $0 < |h| < \delta $. Check that $|\frac{f(x+h) – f(x)}{h} – 3x^2| < \epsilon$.
$$|\frac{f(x+h) – f(x)}{h} – 3x^2|$$
$$= |\frac{x^3 + 3x^2h + 3xh^2 + h^3-x^3}{h} – 3x^2|$$
$$= |\frac{x^3 + 3x^2h + 3xh^2 + h^3-x^3- 3x^2h}{h}|$$
$$= |\frac{3xh^2 + h^3}{h}|$$
$$= |\frac{h(3xh + h^2)}{h}|$$
$$= |3xh + h^2|$$
$$= |h(3x + h)|$$
$$= |h||3x + h|$$
$$< \delta|3x + h|$$
My question is, how do I advance from here? Since there's two variables left, I don't really know how to get rid of the x here (and the {min} strategy wouldn't work either, once again, because of the two variables)- so what should I do now? Is there any way to make a proper epsilon-delta proof for this limit?
Thank you in advance!
Best Answer
What you should do now is not worry about having a slick inequality and just make estimations, regardless of how crude they are. Intuitively for small $h$ we know $3x+h$ is "comparable" to $3x$ which has a "known" size so it should follow that $\delta\cdot|3x+h|$ becomes as small as we please.
How do we formalise "comparable"? Well...
$|3x+h|\le3|x|+|h|$. If $\delta<1$, then $|3x+h|<3|x|+1$. This is the key point; because you have control over the size of $h$, you can basically "remove" it by using this control. It's best to ask yourself why [thing] should be "small" before you try to see how you can prove [thing] is "small".
Now just take any $0<\delta<\min\{1,(3|x|+1)^{-1}\epsilon\}$ and you're happy.