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 $\underset{x\to x_0}{\lim }f(x)=f(x_0).$ $\Rightarrow \underset{x\to x_0}{\lim }f(x)-f(x_0)=0$ $\Rightarrow \underset{x\to x_0}{\lim }f(x)-\underset{x\to x_0}{\lim }f(x_0)=0$ $\Rightarrow \underset{x\to x_0}{\lim }[f(x)-f(x_0)]=0.$ So, to prove the theorem, it is enough to prove that $\underset{x\to x_0}{\lim }[f(x)-f(x_0)]=0.$

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

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

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