**Proof plan.** We prove the special case where $f$ and $g$ are [differentiable](/page/Derivative) at $x_0$ with continuous first derivatives and $f(x_0) = g(x_0) = 0$. The strategy is to expand both $f$ and $g$ via [Taylor's Theorem](/theorems/827) to first order, divide, and cancel the common factor $(x - x_0)$. **Step 1: Apply Taylor's Theorem to numerator and denominator.** Since $f$ and $g$ are differentiable at $x_0$ with $f(x_0) = 0$ and $g(x_0) = 0$, [Taylor's Theorem](/theorems/827) gives: \begin{align*} f(x) &= f(x_0) + (x - x_0) f'(x_0) + o(x - x_0) = (x - x_0) f'(x_0) + o(x - x_0), \\ g(x) &= g(x_0) + (x - x_0) g'(x_0) + o(x - x_0) = (x - x_0) g'(x_0) + o(x - x_0). \end{align*} **Step 2: Form the ratio and cancel.** Dividing both numerator and denominator by $(x - x_0)$: \begin{align*} \frac{f(x)}{g(x)} = \frac{f'(x_0) + o(x - x_0)/(x - x_0)}{g'(x_0) + o(x - x_0)/(x - x_0)}. \end{align*} **Step 3: Take the limit.** As $x \to x_0$, the terms $o(x - x_0)/(x - x_0) \to 0$ by definition. Since $g'(x_0) \neq 0$, the denominator is bounded away from zero for $x$ sufficiently close to $x_0$, and we may evaluate numerator and denominator [limits](/page/Limit) separately: \begin{align*} \lim_{x \to x_0} \frac{f(x)}{g(x)} = \frac{f'(x_0)}{g'(x_0)} = \lim_{x \to x_0} \frac{f'(x)}{g'(x)}, \end{align*} where the final equality uses [continuity](/page/Continuity) of $f'$ and $g'$ at $x_0$.