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!


1 Answer 1


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.

  • $\begingroup$ Small question: when we say 𝛿:=min{1,(3|𝑥|+1)−1𝜖}, can 𝛿 = 1? Since your answer is based off 𝛿 being less than 1, I'm just curious if there is some way to restrict 𝛿 from being equal to 1 in the first place, or if the min{} does that by itself. $\endgroup$ Commented Sep 1, 2023 at 22:03
  • 1
    $\begingroup$ @LogicBeDamned Well, if $\delta=1$ happens then we still have $\delta|3x+h|\color{red}{\le}\epsilon$. It's good to learn not to worry about $\color{red}{\le}\epsilon$ versus $\color{red}{\lt}\epsilon$ (in this context). That said, it was a typo.. should have had $\delta<\min\{1,(3|x|+1)^{-1}\epsilon\}$ (because I know most new students want strict inequalities) $\endgroup$
    – FShrike
    Commented Sep 1, 2023 at 22:23
  • $\begingroup$ You could also do $\delta:=\frac{1}{2}\min\{1,(3|x|+1)^{-1}\epsilon\}$ or $\min\{\frac{1}{2},(3|x|+1)^{-1}\epsilon\}$ or ... there are quite literally uncountably many ways to do this $\endgroup$
    – FShrike
    Commented Sep 1, 2023 at 22:24
  • $\begingroup$ Alright! Gonna bother you with one last question, then, if you don't mind. So, I've set 0 < 𝛿 < min{1,(3|𝑥|+1)−1𝜖}, and as you mentioned in your initial answer, I've gotten to |3𝑥+ℎ |< 3|𝑥|+1. My final question is, how do we then prove that 3|𝑥|+1 ≤ 𝜖? Just couldn't exactly wrap my head around it, so I thought I'd ask just in case. $\endgroup$ Commented Sep 1, 2023 at 22:35
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    $\begingroup$ @LogicBeDamned $3|x|+1\le\epsilon$ won't be true in general. However $\color{red}{\delta\cdot}(3|x|+1)<\epsilon$ because $\delta<(3|x|+1)^{-1}\epsilon$ $\endgroup$
    – FShrike
    Commented Sep 1, 2023 at 22:46

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