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]