Differentiable Function is Continuous

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Theorem

Let $f$ be a real function defined on an interval $I$.

Let $x_0 \in I$ such that $f$ is differentiable at $x_0$.


Then $f$ is continuous at $x_0$.


Corollary

If $f$ is not continuous at $x_0$, $f$ is not differentiable at $x_0$.


Proof

By hypothesis, $f' \left({x_0}\right)$ exists.

We have:

\(\displaystyle f \left({x}\right) - f \left({x_0}\right)\) \(=\) \(\displaystyle \frac {f \left({x}\right) - f \left({x_0}\right)} {x - x_0} \cdot \left({x - x_0}\right)\)
\(\displaystyle \) \(\to\) \(\displaystyle f' \left({x_0}\right) \cdot 0\) as $x \to x_0$


Thus:

$f \left({x}\right) \to f \left({x_0}\right)$ as $x \to x_0$

or in other words:

$\displaystyle \lim_{x \to x_0} \ f \left({x}\right) = f \left({x_0}\right)$


The result follows by definition of continuous.

$\blacksquare$


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