The limit of $f'(x)=b$ implies that $f$ is differentiable at that point [duplicate]

Some hints:

Using the definition of derivative, you need to show that $$ \lim_{h\rightarrow 0^+} {f(a+h)-f(a)\over h } $$ exists and is equal to $\lim\limits_{x\rightarrow a^+} f'(x)$.

Note that for $h>0$ the Mean Value Theorem provides a point $c_h$ with $a<c_h<a+h$ such that $$ {f(a+h)-f(a)\over h } =f'(c_h). $$

Finally, note that $c_h\rightarrow a^+$ as $h\rightarrow0^+$.

The result is essentially Theorem 5.29 from my honors calculus notes. As I mention, I learned this result from Spivak's Calculus. I say "essentially" because the version discussed in my notes is for an interior point of an interval whereas your version is at an endpoint, but to prove the two-sided version you just make a one-sided argument twice, so it's really the same thing.

[And David Mitra is right: the proof uses the Mean Value Theorem and not much else.]

Added: Since we are talking about this result anyway: although I call it a "theorem of Spivak", this is not entirely serious -- the result is presumably much older than Michael Spivak. I am just identifying my (probably secondary) source. If someone knows a primary source, I'd be very happy to hear it.

Also it is of interest to ask what this result is used for. In my notes it isn't used for anything but is only a curiosity. I think Spivak does use it for something, though I forget what at the moment. Moreover a colleague of mine called my attention to this result in the context of, IIRC, Taylor's Theorem. Does anyone know of further applications?

Applying L'Hospital's rule yields: $ \lim_{x \to a} \,(f(x) - f(a)) \mathbin{/} (x - a) = \lim_{x \to a} f'(x) $.

Thus, the assumed existence of $\lim_{x \to a} f'(x)$ proves that the derivative of $f(x)$ at $x = a$ exists and is equal to $\lim_{x \to a} f'(x)$.

You need to show that the limit $$ \lim_{h\to 0^+} \frac{f(a+h)-f(a)}{h} $$ exists and it is equal to $\,L=\lim_{x\to a^+}{f'(x)}$.

Let $\varepsilon>0$. We need to find a $\delta>0$, such that $$ 0<h<\delta\quad\Longrightarrow\quad \left|\,\frac{f(a+h)-f(a)}{h}-L\,\right|<\varepsilon. $$ For this $\,\varepsilon>0,\,$ there exists a $\delta>0$, such that
$$ 0<h<\delta\quad\Longrightarrow\quad \left|\,\,f'(a+h)-L\,\right|<\varepsilon. $$ At the same time, for every $h\in (0,\delta)$, according to the Mean Value Theorem, there exists a $c_h\in (0,1)$, such that $$ \frac{f(a+h)-f(a)}{h}=f(a+\vartheta_h h), $$ and hence $$ \left|\,\frac{f(a+h)-f(a)}{h}-L\,\right|=\left|\,f'(a+\vartheta_h h)-L\,\right| <\varepsilon. $$ The last inequality holds since $0<\vartheta_h h<h<\delta$.