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Theorem 7. Suppose that $f$ is continuous at $a$, and that $f^\prime(x)$ exists for all $x$ in some interval containing $a$, except perhaps for $x = a$. Suppose, moreover, that $lim_{x\to a}f^\prime(x)$ exists. Then $f^\prime(a)$ exists and $f^\prime(a) = lim_{x\to a}f^\prime(x)$

From my understanding, this theorem says that under certain conditions, the derivative of a continuous function is continuous. But, near the end of the proof, I am lost as to how $\lim_{h\to 0}f^\prime(\alpha_h) = \lim_{x\to a} f^\prime(x)$. It seems to me this equality is the crux of the whole proof so can someone break down why the equality holds? Thank you.

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    Possible to provide the theorem and the proof? – 2017-01-07
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    I think the proof of the equality is obvious because as $h\to 0$, the quantity $\alpha_{h} \to a$. It can be rigorously proven using $\epsilon, \delta$. – 2017-01-07

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