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Let $f$ differentiable at $x_0$. Show that the following limit exists

$$ \lim_{h\rightarrow0} \frac{f(x_0+h)-f(x_0-h)}{h}$$

If $f$ is differetiable at $x_0$ then it's one-sided derivative exists and equal. Hence,

$$ \lim_{h\rightarrow0^+} \frac{f(x_0 +h)-f(x_0)}{h} = \lim_{h\rightarrow0^-} \frac{f(x_0 +h)-f(x_0)}{h} $$

Now, technically if I do a simple arithmetic I can get the answer (move the right limit and "join" them). Moreover, the limit exists and equals $0$.

But, I cannot just join them because they're not the same.

What should I do?

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  • $\begingroup$ $$\lim_{h\rightarrow 0}\dfrac{f(x_0+h)-f(x_0)}{h}$$ exists also implies that $$\lim_{h\rightarrow 0}\dfrac{f(x_0-h)-f(x_0)}{h}$$ exists. So I don't think there should be any problem. $\endgroup$ – Debashish Jun 19 '14 at 9:19
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You can do the following:

\begin{align*} \lim_{h\rightarrow0} \frac{f(x_0+h)-f(x_0-h)}{h} &= \lim_{h\rightarrow0} \frac{f(x_0+h)-f(x_0)+f(x_0)-f(x_0-h)}{h}\\ & = \lim_{h\rightarrow0} \frac{f(x_0+h)-f(x_0)}{h}+\lim_{h \to 0}\frac{f(x_0)-f(x_0-h)}{h}\\ \end{align*} you can split this because both these limits exist as $f$ is differentiable at $x_0$. If you have doubts about the second limit, then just think of $h=-t$ and it becomes $$\lim_{t \to 0}\frac{f(x_0)-f(x_0+t)}{-t}.$$

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  • 1
    $\begingroup$ You should not take $x_0-h=t$, otherwise $t$ will depend on $h$, which is a problem taking limit. Rather think of $\tilde{h} = -h$, then you get $\frac{f(x_0)-f(x_0-h)}{h}= \frac{f(x_0)-f(x_0+\tilde{h})}{-\tilde{h}} = \frac{f(x_0+\tilde{h})-f(x_0)}{\tilde{h}}$ $\endgroup$ – mlk Jun 19 '14 at 9:29
  • $\begingroup$ @mlk what you point out is right but $t+h$ remains a constant so we can still justify it. $\endgroup$ – Anurag A Jun 19 '14 at 9:38
  • $\begingroup$ Your answer is correct but as pointed by Anurag A what you mention at the end is wrong. The right substitution is $h = -t$ which transforms it to $\{f(x_{0} + t) - f(x_{0})\}/t$ and $t \to 0$ $\endgroup$ – Paramanand Singh Jun 20 '14 at 13:46

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