L'Hôpital's Rule

L’Hôpital’s Rule (0/0 form, one-sided): Let $a<b$, and let $f,g:[a,b)\to\mathbb{R}$ be continuous on $[a,b)$ and differentiable on $(a,b)$. Assume:

• $f(a)=g(a)=0$,
• $g'(x)\neq 0$ for all $x\in(a,b)$,
• the limit $L=\lim_{x\to a^+}\frac{f'(x)}{g'(x)}$ exists in $\mathbb{R}\cup\{\pm\infty\}$.

Then the limit $\lim_{x\to a^+}\frac{f(x)}{g(x)}$ exists and equals $L$: $\lim_{x\to a^+}\frac{f(x)}{g(x)}=\lim_{x\to a^+}\frac{f'(x)}{g'(x)}=L.$

This rule is a standard tool for evaluating difficult limits, but it must be used with all hypotheses in place (especially the differentiability and nonvanishing of $g'$ near the limit point).

Proof sketch: For $x\in(a,b)$, apply the Cauchy mean value theorem to $f$ and $g$ on $[a,x]$. There exists $c\in(a,x)$ such that $\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f'(c)}{g'(c)}.$ Since $f(a)=g(a)=0$, this becomes $\frac{f(x)}{g(x)}=\frac{f'(c)}{g'(c)}$. As $x\to a^+$ we have $c\to a^+$, and the right-hand side tends to $L$.