# Differentiable Function is Continuous

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## Contents

## 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.

$\blacksquare$

## Sources

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 10.6$