Differentiability Theorem
If \(f\) is differentiable at a point \(x = a\), then \(f\) must be continuous at \(x = a\)

Proof

Since \(f\) is differentiable at \(x = a\), then by defintion

$$ \begin{align*} f'(a) = \lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a} \end{align*} $$

exists. Now note that

$$ \begin{align*} f(x) &= f(x) \\ f(x) &= (f(x) - f(a)) + f(a) \\ f(x) &= \frac{f(x) - f(a)}{x - a} \cdot (x - a) + f(a) \end{align*} $$

Applying the limit on both sides

$$ \begin{align*} \lim\limits_{x \to a} f(x) &= \lim\limits_{x \to a} \left[ \frac{f(x) - f(a)}{x - a} \cdot (x - a) + f(a) \right] \\ \lim\limits_{x \to a} f(x) &= \lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a} \cdot \lim\limits_{x \to a} (x - a) + \lim\limits_{x \to a} f(a) \\ \lim\limits_{x \to a} f(x) &= f'(a) \cdot 0 + f(a) \\ \lim\limits_{x \to a} f(x) &= f(a) \end{align*} $$

Hence, \(f\) is continuous at \(x = a\). \(\ \blacksquare\)
Note: the converse is false!

References