Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $f$ is differentiable at $k$, find: $\lim_{h \to 0} \frac{f(k + ph) - f(k - ph)}{h}$

I realize that since the limit exists at k, then:

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

and I can visualize what might be happening on the coordinate axis: the two point on each axis are getting further apart it seems?

But I'm not sure how this has all affected the limit in the question.

share|cite|improve this question
For those who might be interested, variations on this notion of differentiation are called the pseudo-symmetric differentiation and parametric differentiation. See the following posts for references to some research papers on these notions: (e.g. How large can the set of points of non-differentiability be for an everywhere pseudo-symmetrically differentiable function?) and – Dave L. Renfro Apr 13 '12 at 14:44
up vote 1 down vote accepted

for $p = 0$, your limit is $0$. Now suppose $p \ne 0$. We have \begin{align*} \lim_{h \to 0} \frac{f(k+ph) - f(k-ph)}h &= \lim_{h\to 0}\left( \frac{f(k+ph)-f(k)}h - \frac{f(k-ph) - f(k)}h\right)\\\ &= \lim_{h\to 0} \left( p\frac{f(k+ph) - f(k)}{ph} + p\frac{f(k-ph) - f(k)}{-ph}\right)\\\ &= p \lim_{\eta \to 0} \frac{f(k+\eta) - f(k)}\eta + p \lim_{\eta\to 0}\frac{f(k + \eta) - f(k)}\eta\\\ &= 2pf'(k). \end{align*}

Hope this helps.

share|cite|improve this answer
I don't understand how $ph$ and $-ph$ both became $\eta$? Or is it just an interpretation since both are the same distance from k? – stariz77 Apr 13 '12 at 8:27

$$\frac{f(k+ph)-f(k-ph)}h=p\cdot\left(\frac{f(k+ph)-f(k)}{ph}-\frac{f(k-ph)-f(k)}{ph}\right) $$

share|cite|improve this answer

May be try putting $t =ph$. Then as $h \to 0, \ ph \to 0 \Rightarrow t \to 0$. So your limit is nothing but $$\lim_{t \to 0} p \cdot \frac{f(k+t) - f(k-t)}{t}$$ I guess you can evaluate this now.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.