Proof that differentiable functions are continuous

Proof

1. Since $f$ is differentiable at $a$, it has a derivative $L = f'(a)$. For $h \ne 0$ with $a + h \in I$, write $q(h) = \frac{f(a + h) - f(a)}{h}$. Then $q(h) \to L$ as $h \to 0$.
2. By the epsilon-delta form of limits with $\varepsilon = 1$, there is $\delta_1 > 0$ with $\abs{q(h) - L} < 1$ whenever $0 < \abs{h} < \delta_1$ and $a + h \in I$.
3. For such $h$, the triangle inequality gives $\abs{q(h)} \le \abs{q(h) - L} + \abs{L} < \abs{L} + 1$.
4. So $\abs{f(a + h) - f(a)} = \abs{q(h)} \, \abs{h} \le (\abs{L} + 1) \abs{h}$ for these $h$. This also holds for $h = 0$, where both sides are $0$.
5. Let $\varepsilon > 0$. Put $\delta = \min\left(\delta_1, \frac{\varepsilon}{\abs{L} + 1}\right) > 0$.
6. Let $x \in I$ with $\abs{x - a} < \delta$, and put $h = x - a$. By step 4, $\abs{f(x) - f(a)} \le (\abs{L} + 1) \abs{h} < (\abs{L} + 1) \cdot \frac{\varepsilon}{\abs{L} + 1} = \varepsilon$.
7. By the epsilon-delta form of continuity, $f$ is continuous at $a$.