Differentiability Implies Continuity

Theorem

Let $f : I \to R$ where $I$ is an interval. If $f$ is differentiable at $c \in I$ , then $f$ is continuous at $c$ .

Proof

For all $x \in I$ where $x \neq c$ then we have the following:

\begin{aligned} f (x) - f (c) & = \frac{f (x) - f (c)}{x - c} \cdot (x - c) \\ lim_{x \to c} [f (x) - f (c)] & = lim_{x \to c} [\frac{f (x) - f (c)}{x - c} \cdot (x - c)] \\ lim_{x \to c} f (x) - f (c) & = (lim_{x \to c} \frac{f (x) - f (c)}{x - c}) (lim_{x \to c} (x - c)) \\ = f^{'} (c) \cdot 0 & ALT \\ lim_{x \to c} [f (x) - f (c)] & = 0 \end{aligned}

Then if we want to compute the limit of $f (x)$ at $c$ can be calculated:

\begin{aligned} lim_{x \to c} f (x) & = lim_{x \to c} (f (x) - f (c)) + f (c) \\ = lim_{x \to c} f (x) - f (c) + lim_{x \to c} f (c) \\ = 0 + f (c) \\ = f (c) \end{aligned}

Thus $lim_{x \to c} f (x) = f (c)$ so the limit exists and is as we expect.

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