I am studying a proof of a simple l'Hôpital's rule.

The theorem is as follows:

Let $x_0 \in (a, b).$
Suppose that $f, g$ are differentiable on $(a, b)$.
Suppose $\lim_{x\to x_0}f(x) = \lim_{x\to x_0}g(x) = 0$ and $g'(x) \not= 0.$
Then $$\lim_{x\to x_0}\frac{f(x)}{g(x)} = \frac{f'(x_0)}{g'(x_0)}.$$

The beginning of the proof is as follows:

Since $g'(x_0) \not= 0$, there is an interval $(x_0 - r, x_0 + r) \subset (a, b)$ on which $g(x) - g(x_0) \not= 0.$
By the algebra of limits, for $x \in (x_0 - r, x_0 + r),$ $$\lim_{x\to x_0}\frac{f(x)}{g(x)} = \lim_{x\to x_0}\frac{\frac{f(x) - f(x_0)}{x - x_0}}{\frac{g(x) - g(x_0)}{x - x_0}} = \cdots$$

NB: I have cut the proof short by inserting the ellipsis.
I do understand why both $f(x)$ and $g(x)$ can be divided by $x - x_0$.
However, I do not yet understand why $$\lim_{x\to x_0}\frac{f(x)}{g(x)} = \lim_{x\to x_0}\frac{f(x) - f(x_0)}{g(x) - g(x_0)}.$$ This would be the implicit step taken before dividing through by $x - x_0$.
Could someone please explain why this implicit step holds?


Because by assumption $f(x_0)=g(x_0)$. Well actually only $\lim_{x\to x_0}f(x)=\lim_{x\to x_0}g(x)=0$, gut as $f,g$ are differentiable at $x_0$, they are also continuous at $x_0$.

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  • $\begingroup$ it is important that $f(x_0)=g(x_0)=0$ $\endgroup$ – Bananach May 28 '15 at 15:14

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