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

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.

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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?) mathforum.org/kb/thread.jspa?messageID=688881 and mathforum.org/kb/message.jspa?messageID=688882 – Dave L. Renfro Apr 13 '12 at 14:44

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.

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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

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

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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.

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