Suppose that $f : [a,b] \rightarrow \mathbb R$ is continuous on $[a,b]$, differentiable on $(a,b)$, and that $\lim_{x\rightarrow a^+}f'(x)=L$. Show that $f$ is differentiable at $a$, and that $f'(a) = L$.
I have tried starting with the continuity of the function, but I'm still not sure how to even begin or where to go from there.
Not a hard problem. I say that because the hypotheses are very close to the conclusion ... provided, however, it occurs to you to use the mean-value theorem. The only other reason why you couldn't handle it is because you need more experience working with limits. In any case after you write up a solution do reflect for a while on why it caused some difficulty! A two-step problem shouldn't. No shame, but you do need to figure out why routine problems are causing you any grief. Things do get more complicated later on.
Here the pieces of the puzzle that merely need to be assembled:
- The definition of $f'(a)$ is $$f'(a) = \lim_{x\to a+} \frac{f(x)-f(a)}{x-a},$$ this because it is only the right-hand derivative at $a$ that can be considered.
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- The mean value theorem says that $\frac{f(x)-f(a)}{x-a}= f'(\xi)$ for some value $\xi$ between $a$ and $x$.
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- The hypothesis that $\lim_{x\rightarrow a^+}f'(x)=L$ says that $f'(\xi)$ is as close as you please to $L$ if $\xi$ is close enough to $a$.
Another poster suggested the mean-value theorem. Let's think about why. The connection between a function $f$ and its derivative $f'$ in one direction (i.e., obtaining information about $f'$ from $f$) is the definition itself; and in the other direction (i.e., obtaining information about $f$ from $f'$) is, for calculus students, the mean-value theorem. There are more sophisticated tools that you will learn later on, but for novices the mean-value theorem should always come immediately to mind.)
