Does the existence of $\lim f'(x)$ imply the existence of $\lim f(x)$? Let $f(x)$ be a differentiable function on an interval $(a,b)$, where $a<b$. If $\lim_{x\to a^+}f'(x)$ exists, does it necessarily follow that $\lim_{x\to a^+}f(x)$ also exists? I suspect that it is true, but does anyone know any simple proof?
 A: By MVT $f(x)-f(y)=(x-y)f'(t)$ for some $t$ between $x$ and $a$. Since $f'(t)$ remains bounded for $x,y>a, |x-a|,|y-a|$ sufficiently small it follows that $(f(x_n))$ is a Cauchy sequence for any $x_n$ decreasing to $a$. hence it has a limit. It is also clear form MVT that the limit is independent of $(x_n)$. hence $\lim_{x \to a+} f(x)$ exists.
A: Edit: I forgot differentiable and $C^1$ were distinct here for a moment. This answer assumes that $f$ is $C^1$ (at least on some interval $(a,a+2\epsilon)$).
Further edit: Actually, rereading the statement, FTC doesn't require $f'$ to be continuous, merely Riemann integrable. Thus we just need $f'$ to be Riemann integrable in some open set $(a,a+2\epsilon)$. This is still not generally true if $f$ is differentiable, since there are functions like the Volterra function, which are differentiable everywhere with bounded derivative, but whose discontinuities have positive measure.
Original answer
By the fundamental theorem of calculus, $$f(x)=f(a+\epsilon) - \int_x^{a+\epsilon}f'(t)\,dt,$$ and as $x\to a$, $$f(x)\to f(a+\epsilon) - \int_a^{a+\epsilon}f'(t)\,dt.$$ The integral exists, and the limit converges since if $$\lim_{x\to a^+}f'(x)$$
exists, then $f'(x)$ is bounded on $(a,a+\epsilon)$ for some $\epsilon > 0$.
As a side note, we only need that $f'$ is bounded on $(a,a+\epsilon)$ for some $\epsilon$ for this proof to work.
A: Suppose $\lim_{x \rightarrow a^+}f'(x)=L$. Given $\epsilon > 0$, there is a $\delta > 0$ such that $a < x < a + \delta$ implies $L - \epsilon < f'(x) < L + \epsilon$. Then, if $a < x < y < a + \delta$, we have, by an application of the MVT, $L - \epsilon < \frac{f(y)-f(x)}{y-x} < L + \epsilon$, i.e., 
$$(*): f(x)+(L-\epsilon)(y-x) < f(y) < f(x)+(L+\epsilon)(y-x)$$
It follows that $f(y) \le (\liminf_{x \rightarrow a^+} f(x)) + (L+\epsilon)(y-a)$ and therefore $$\limsup_{y \rightarrow a^+} f(y) \le \liminf_{x \rightarrow a^+} f(x)$$
By (*), it follows that $f$ is bounded near $a$ and then $\lim_{x \rightarrow a^+} f(x)$ exists.
