# differentiable functions are continuous

###### Proposition 1.

Suppose $I$ is an open interval on $\mathrm{R}$,
and $f\mathrm{:}I\mathrm{\to}\mathrm{C}$ is differentiable^{} at $x\mathrm{\in}I$. Then
$f$ is continuous at $x$. Further, if $f$ is differentiable on $I$,
then $f$ is continuous on $I$.

###### Proof.

Suppose $x\in I$. Let us show that $f(y)\to f(x)$, when $y\to x$. First, if $y\in I$ is distinct to $x$, then

$$f(x)-f(y)=\frac{f(x)-f(y)}{x-y}(x-y).$$ |

Thus, if ${f}^{\prime}(x)$ is the derivative of $f$ at $x$, we have

$\underset{y\to x}{lim}f(x)-f(y)$ | $=$ | $\underset{y\to x}{lim}{\displaystyle \frac{f(x)-f(y)}{x-y}}(x-y)$ | ||

$=$ | $\underset{y\to x}{lim}{\displaystyle \frac{f(x)-f(y)}{x-y}}\underset{y\to x}{lim}(x-y)$ | |||

$=$ | ${f}^{\prime}(x)\mathrm{\hspace{0.25em}0}$ | |||

$=$ | $0,$ |

where the second equality is justified since both limits on the second line exist. The second claim follows since $f$ is continuous on $I$ if and only if $f$ is continuous at $x$ for all $x\in I$. ∎

Title | differentiable functions are continuous |
---|---|

Canonical name | DifferentiableFunctionsAreContinuous |

Date of creation | 2013-03-22 14:35:27 |

Last modified on | 2013-03-22 14:35:27 |

Owner | matte (1858) |

Last modified by | matte (1858) |

Numerical id | 8 |

Author | matte (1858) |

Entry type | Theorem |

Classification | msc 57R35 |

Classification | msc 26A24 |

Related topic | DifferentiableFunctionsAreContinuous2 |

Related topic | LimitsOfNaturalLogarithm |