**Problem Statement: **Suppose that is differentiable on the bounded interval . Suppose further that there is a constant such that for every . Show that exists and is finite.

**Note:** When I read this question the first time I thought the qual writers were crazy because I found a counter example. If we have a piecewise smooth function on that is continuous on with a jump discontinuity for then the doesn’t have to exist since the right and left handed limits are not equal. Done! But, after much deliberation I’ve concluded that the qual writers are not crazy and they must really be asking me to show that exists and is finite. So, with that in mind I was able to continue.

**Proof:**

Since is differentiable on it follows that is continuous on . Let such that for every .

*Claim:* is uniformly continuous on . Let and wlog . Then is continuous and differentiable on , so we may apply the Mean Value Theorem, which states that there exists a such that . By our assumption it follows that . This implies that . Let and . Then for it follows that . Since x and y were arbitrary in it follows that is uniformly continuous on .

Let be a sequence converging to such that for every n. Since this is a convergent sequence we get that must be Cauchy. Let where is as defined above. Then since is Cauchy there exists and such that whenever . Since this implies that . Thus, is Cauchy and so converges as . Therefore exists and is finite. But this is equivilent to the statement that the exists and is finite.

This is a good problem, it uses two of my favorite things from analysis: mean value theorem and Cauchy sequences! Bounded derivative implies uniform continuity, that’s a classic application of mvt.

I think it’s not done yet! You have shown that the image sequence of any Cauchy sequence in the domain that tends to a converges to something. You must argue that they _all_ tend to the same limit L. That common value is then the limit of f(x) as x tends to a.

Keep up the good work…

So, I think right before the second to last sentence I should add “Let . Suppose is another sequence in convering to . Then and must converge to the same limit, L, as since limits are unique. ”

Is that enough to say or should I go into that argument in more detail?

Thanks for the feedback!

I want to note that we need f(x) to be uniformly continuous on (a,b) so that we can obtain the inequality, when , so that our sequence is Cauchy and therefore convergent.