1. ## Differentiation (Real Analysis)

f:[a,b]->R is continuous on [a,b] and differentiable on (a,b). Show $\lim_{x \rightarrow\ a}f'(x)=A$, then f'(a) exists and equals A.

The hint we got was to use the definition of f'(a) and the Mean Value theorem, but I"m still having no luck. Can someone please help?

2. Suppose that $\varepsilon > 0$ because $\lim _{x \to a} f'(x) = A$, we know that $\left( {\exists \delta > 0} \right)\left[ {\left| {a - x} \right| < \delta \Rightarrow \left| {f'(x) - A} \right| < \varepsilon } \right]$.
By the mean value theorem: $a < x < a+\delta \Rightarrow \left( {\exists c_x } \right)\left[ {a < c_x < x \wedge f'\left( {c_x } \right) = \frac{{f(x) - f(a)}}{{x - a}}} \right]$.
Now look at this:
$\left| {\frac{{f(x) - f(a)}}{{x - a}} - A} \right| = \left| {f'\left( {c_x } \right) - A} \right| < \varepsilon \mbox{ because } \left| {a - c_x } \right| < \left| {a - x} \right| < \delta$

3. NM, dumb post.

4. Originally Posted by hockey777
NM, dumb post.
Please tell me what that non-standard English expression means?