Differentiation (Real Analysis)

**hockey777** · October 30th 2008, 01:49 PM

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?

**Plato** · October 30th 2008, 02:43 PM

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:
$\left| {\frac{{f(x) - f(a)}}{{x - a}} - A} \right| = \left| {f'\left( {c_x } \right) - A} \right| < \varepsilon \mbox{ because } \left| {a - c_x } \right| < \left| {a - x} \right| < \delta$

**hockey777** · October 30th 2008, 04:57 PM

NM, dumb post.

**Plato** · October 30th 2008, 05:32 PM

Originally Posted by hockey777

NM, dumb post.

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

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