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I had a question about an answer I saw on this website. Here's the link to the full question/answer correspondence (I've also copied the relevant text below):

Second derivative "formula derivation"

This is my question: I've never seen limits like "as -k approaches zero." When you multiply a limit by negative number (or simply factor out a negative from the function that you are taking the limit of) does that mean you always flip the sign of the number that approaches zero? For example, is this true?

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

I think I saw something very similar, in the answer below (if I understood it correctly). BTW--thanks to the author, Brian M. Scott, for contributing this very helpful answer!

The only problem is that you’re looking at the wrong three points: you’re looking at $x+2h,x+h$, and $x$, and the version that you want to prove is using $x+h,x$, and $x-h$. Start with $$f''(x)=\lim_{h\to 0}\frac{f'(x)-f'(x-h)}h,$$ and you’ll be fine.

To see that this really is equivalent to looking at $$f''(x)=\lim_{h\to 0}\frac{f'(x+h)-f'(x)}h,$$ let $k=-h$; then

$$\begin{align*} f''(x)&=\lim_{h\to 0}\frac{f'(x)-f'(x-h)}h\\ &=\lim_{\!\!-k\to0\!\!}\frac{f'(x)-f'(x-(-k))}{-k}\\ &=\lim_{k\to 0}\frac{f'(x-(-k))-f(x)}k\\ &=\lim_{k\to 0}\frac{f'(x+k)-f(x)}k, \end{align*}$$

and renaming the dummy variable back to $h$ completes the demonstration.

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Derek
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1 Answers1

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By definition, $\lim_{h \to 0} f(h) = l$ means that $\forall \epsilon > 0, \exists \delta > 0$ such that $0 < |h| < \delta \implies |f(h)-l| < \epsilon$. Thus it's actually equivalent to use a new variable $k = -h$ and to ask that $\forall \epsilon > 0, \exists \delta > 0$ such that $0 < |k| < \delta \implies |f(-k)-l| < \epsilon$.

Alex Provost
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