# Timeless Theorems of Mathematics/Differentiability Implies Continuity

**Statement:** If a function $f(x)$ is differentiable at the point
$x_0,$ then $f$ is continuous at $x_0.$

**Proof:** Assume $f(x)$ is differentiable at $x_0$. Then, $D_{x}f(x_0)$
exists.

The function $f(x)$ is continuous at $x_0$ when
$\lim_{x \to x_0} f(x) = f(x_0).$
$\Rightarrow \lim_{x \to x_0} f(x) - f(x_0) = 0$
$\Rightarrow \lim_{x \to x_0} f(x) - \lim_{x \to x_0} f(x_0) = 0$
$\Rightarrow \lim_{x \to x_0} [f(x) - f(x_0)] = 0.$ So, to prove the
theorem, it is enough to prove that
$\lim_{x \to x_0} [f(x) - f(x_0)] = 0.$

$\lim_{x \to x_0} [f(x) - f(x_0)]$
$=\lim_{x \to x_0} [\frac {f(x) - f(x_0)}{x - x_0} (x-x_0)]$
$=\lim_{x \to x_0} [\frac {f(x) - f(x_0)}{x - x_0}]\cdot\lim_{x \to x_0} (x-x_0)$
$=D_{x}f(x_0)\cdot 0 = 0$

Therefore, When $D_{x}f(x_0)$ exists,
$\lim_{x \to x_0} [f(x) - f(x_0)] = 0$ is true.

$\therefore$ If $f(x)$ is differentiable at the point $x_0,$ then $f$ is
continuous at $x_0.$ ***\[Proved\]***

