[proofplan] We show differentiability implies continuity by writing $f(a+h) - f(a)$ as the product of $h$ and the difference quotient, then using the product rule for limits. Since $h \to 0$ and the difference quotient converges to $f'(a)$, the product tends to $0 \cdot f'(a) = 0$, giving $f(a+h) \to f(a)$. [/proofplan] [step:Factor the function increment as a product of $h$ and the difference quotient] Since $f$ is differentiable at $a$, there exists $f'(a) \in \mathbb{R}$ such that \begin{align*} \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = f'(a) \end{align*} where $h$ ranges over non-zero values with $a + h \in E$. For such $h$, we write \begin{align*} f(a+h) - f(a) = h \cdot \frac{f(a+h) - f(a)}{h}. \end{align*} [/step] [step:Apply the product rule for limits to conclude $f(a+h) - f(a) \to 0$] Taking the limit as $h \to 0$ and applying the product rule for limits (since both limits exist): \begin{align*} \lim_{h \to 0} \bigl(f(a+h) - f(a)\bigr) &= \lim_{h \to 0} h \cdot \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \\ &= 0 \cdot f'(a) = 0. \end{align*} Therefore $\lim_{h \to 0} f(a+h) = f(a)$, which is the definition of continuity of $f$ at $a$. [guided] The proof rests on a simple algebraic identity: $f(a+h) - f(a) = h \cdot \frac{f(a+h) - f(a)}{h}$. This is valid for all $h \neq 0$ with $a + h \in E$. To take the limit as $h \to 0$, we use the [product rule for limits](/theorems/170) (part (v)). The first factor $h$ tends to $0$, and the second factor -- the difference quotient -- tends to $f'(a)$ by the definition of differentiability. Both limits exist and are finite, so the product rule applies: \begin{align*} \lim_{h \to 0} \bigl(f(a+h) - f(a)\bigr) = 0 \cdot f'(a) = 0. \end{align*} This gives $\lim_{h \to 0} f(a+h) = f(a)$, which is precisely the statement that $f$ is continuous at $a$. Note that the converse fails: $f(x) = |x|$ is continuous at $0$ but not differentiable there. [/guided] [/step]