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]