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This question already has an answer here:

Suppose that $h$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and that $c \in (a, b)$. Suppose also that $\lim\limits_{x \to c} h'(x)$ exists. Prove that $h'$ is continuous at $c$.

I really have no idea how to think about this problem. I know if the limit exists then it's differentiable at the point, i was assuming differentiability implied continuity. Someone please give me advice.

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marked as duplicate by Hans Lundmark, Jonas Dahlbæk, user370967, Frits Veerman, kingW3 Jun 29 '17 at 13:49

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  • $\begingroup$ Yes, differentiability of $h$ implies continuity of $h$. But you're asked to prove the continuity of $h'$ at $c$, not of $h$. $\endgroup$ – user49640 Apr 6 '17 at 2:57
  • $\begingroup$ |x-c|<$\delta$$\implies$ |h'(x)-h'(c)|<$\epsilon$ $\endgroup$ – August Haze Apr 6 '17 at 3:04
  • $\begingroup$ Since you're assuming that $\lim_{x \to c} h'(x)$ exists, why don't you give it a name? Then assume it's different from $h'(c)$ and try to derive a contradiction. You should start by drawing a picture to try to see why this situation isn't possible. Bear in mind the mean-value theorem. $\endgroup$ – user49640 Apr 6 '17 at 3:19
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To expand on user49640's comment, by the MVT $$\frac{h(c + \delta) - h(c)}{\delta} = h'(\xi(\delta)).$$Since $c - |\delta| \le\xi(\delta) \le c + |\delta|$ we have that $\xi(\delta) \to c$ as $\delta \to 0$. Let $L = \lim_{x \to c}h'(x)$. Then $$h'(c) = \lim_{\delta \to 0}\frac{h(c + \delta) - h(c)}{\delta} = \lim_{\delta \to 0}h'(\xi(\delta)) = L.$$

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Let $f$ be defined as

$$f(x,y)= \begin{cases} \frac{h(x+y)-h(x)}{y}&, y\ne 0\\\\ h'(x)&,y=0 \end{cases}$$

for $x\in (a,b)$ and $x+y\in (a,b)$.

Since $h$ is continuous on $[a,b]$, then for $y\ne 0$, $f(x,y)$ converges uniformly to $f(c,y)$ as $x\to c$.

In addition $\lim_{y\to 0}f(x,y)=h'(x)$.

Then, the Moore-Osgood Theorem guarantees that

$$\lim_{x\to c}\lim_{y\to 0}f(x,y)=\lim_{y\to 0}\lim_{x\to c}f(x,y)\tag 1$$

The left-hand side of $(1)$ is equal to $\lim_{x\to c}h'(x)$. The right-hand side of $(1)$ is equal to $h'(c)$.

And we are done!

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  • $\begingroup$ This is beyond the scope of my class $\endgroup$ – August Haze Apr 6 '17 at 3:33
  • $\begingroup$ @AugustHaze Have you discussed uniform convergence? $\endgroup$ – Mark Viola Apr 6 '17 at 3:33
  • $\begingroup$ We did discuss uniform convergence briefly because we were in a time crunch. $\endgroup$ – August Haze Apr 6 '17 at 3:36
  • $\begingroup$ Well, this solution is within the scope of the class. The thrust of this way forward is the justification of interchanging the order of the limits. $\endgroup$ – Mark Viola Apr 6 '17 at 3:39
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This is a standard property of derivatives: derivatives can't have jump discontinuity. In other words if both $f'(c)$ and $\lim_{x \to c}f'(x)$ exist then they are equal and hence $f'$ is continuous at $c$.

To prove the above property we can use either Mean Value Theorem or Darboux Theorem. First let's use the mean value theorem. We have \begin{align} f'(c) &= \lim_{h \to 0}\frac{f(c + h) - f(c)}{h}\notag\\ &= \lim_{h \to 0}f'(c_h)\text{ (for some }c_h\text{ between }c, c + h)\notag\\ &= \lim_{x \to c}f'(x)\notag\\ \end{align}

Next we use Darboux theorem which says that derivatives possess intermediate value property. Let $f'(c) = A \neq B = \lim_{x \to c}f'(x)$. Let's take the case when $A < B$. By choosing $\epsilon = (B - A)/2$ and using limit $\lim_{x \to c}f'(x) = B$ we can ensure that there is an $h > 0$ such that $$f'(x) > B - \epsilon = A + \epsilon\tag{1}$$ for all $x \in (c - h, c + h)$. Thus in the interval $(c - h, c + h)$ there is a value of $f'$ less than $A + \epsilon$ namely $f'(c)$ and also as noted above there are values of $f'$ greater than $A + \epsilon$. Hence by intermediate value property $f'$ must take the value $A + \epsilon$ in this interval. But this is not possible because of equation $(1)$ and hence we reach a contradiction. If $A > B$ then we can just interchange the roles of $A, B$ in above argument. Hence we must have $A = B$ and then $f'$ is continuous at $c$.

Note: I have used letter $f$ for the function instead of $h$ as given in question. When dealing with just one function $f$ appears to be a natural choice and I keep the answer that way. I wonder why the question uses $h$ unless it is a part of a larger question and $f, g$ have been used up in other parts of the question.

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  • $\begingroup$ Paramanand, hope you're doing well. The MVT certainly applies, but the Moore-Osgood Theorem seemed more general in that it gives sufficient conditions on interchanging limits. -Mark $\endgroup$ – Mark Viola Apr 6 '17 at 22:38
  • $\begingroup$ @Dr.MV: thinking that this was a well known property of Derivatives I did not look at other answers. Now that you mention about your approach I have the opportunity to learn about a new theorem. Thanks and +1 for your answer. $\endgroup$ – Paramanand Singh Apr 7 '17 at 5:50
  • $\begingroup$ You're welcome. Pleased to hear it was useful! $\endgroup$ – Mark Viola Apr 7 '17 at 13:50

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