# 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, $\map {f'} {x_0}$ exists.

We have:

\(\ds \map f x - \map f {x_0}\) | \(=\) | \(\ds \frac {\map f x - \map f {x_0} } {x - x_0} \cdot \paren {x - x_0}\) | ||||||||||||

\(\ds \) | \(\to\) | \(\ds \map {f'} {x_0} \cdot 0\) | as $x \to x_0$ |

Thus:

- $\map f x \to \map f {x_0}$ as $x \to x_0$

or in other words:

- $\ds \lim_{x \mathop \to x_0} \map f x = \map f {x_0}$

The result follows by definition of continuous.

$\blacksquare$

## Also see

## Sources

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