# Differentiation (Real Analysis)

• Oct 30th 2008, 01:49 PM
hockey777
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?
• Oct 30th 2008, 02:43 PM
Plato
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:
$\left| {\frac{{f(x) - f(a)}}{{x - a}} - A} \right| = \left| {f'\left( {c_x } \right) - A} \right| < \varepsilon \mbox{ because } \left| {a - c_x } \right| < \left| {a - x} \right| < \delta$
• Oct 30th 2008, 04:57 PM
hockey777
NM, dumb post.
• Oct 30th 2008, 05:32 PM
Plato
Quote:

Originally Posted by hockey777
NM, dumb post.

Please tell me what that non-standard English expression means?