# Continuity of the derivative at a point given certain hypotheses [duplicate]

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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.

## marked as duplicate by Hans Lundmark, Jonas Dahlbæk, user370967, Frits Veerman, kingW3Jun 29 '17 at 13:49

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• Yes, differentiability of $h$ implies continuity of $h$. But you're asked to prove the continuity of $h'$ at $c$, not of $h$. – user49640 Apr 6 '17 at 2:57
• |x-c|<\delta$$\implies |h'(x)-h'(c)|<\epsilon – August Haze Apr 6 '17 at 3:04 • Since you're assuming that \lim_{x \to c} h'(x) exists, why don't you give it a name? Then assume it's different from h'(c) and try to derive a contradiction. You should start by drawing a picture to try to see why this situation isn't possible. Bear in mind the mean-value theorem. – user49640 Apr 6 '17 at 3:19 ## 3 Answers To expand on user49640's comment, by the MVT$$\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.$$Let f be defined as$$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 1The 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! • This is beyond the scope of my class – August Haze Apr 6 '17 at 3:33 • @AugustHaze Have you discussed uniform convergence? – Mark Viola Apr 6 '17 at 3:33 • We did discuss uniform convergence briefly because we were in a time crunch. – August Haze Apr 6 '17 at 3:36 • 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. – Mark Viola Apr 6 '17 at 3:39 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 thatf'(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.

• 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 – Mark Viola Apr 6 '17 at 22:38
• @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. – Paramanand Singh Apr 7 '17 at 5:50
• You're welcome. Pleased to hear it was useful! – Mark Viola Apr 7 '17 at 13:50