Does the existence of $\lim f'(x)$ imply the existence of $\lim f(x)$?

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 in dependent of $(x_n)$. hence $\lim_{x \to a+} f(x)$ exists.

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.