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TheoremProved

L'Hôpital's rule

Statement

Suppose f,gf,g are differentiable near aa (except possibly at aa), g′(x)≠0g'(x)\neq 0 nearby, and either lim⁡x→af(x)=lim⁡x→ag(x)=0\lim_{x\to a} f(x)=\lim_{x\to a}g(x)=0 or both tend to ±∞\pm\infty. If lim⁡x→af′(x)g′(x)\lim_{x\to a}\frac{f'(x)}{g'(x)} exists (or is ±∞\pm\infty), then lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}.

Why is it true?

Near a tied '0/0' race between two quantities both shrinking to nothing, it's not the sizes but the relative speeds at which they vanish that decide the limit of their ratio - whichever one is shrinking 'faster' (as measured by derivatives) dominates.

Proof sketch

Extend f,gf,g by setting f(a)=g(a)=0f(a)=g(a)=0 (in the 0/00/0 case), so both are continuous at aa. For xx near aa, apply the Cauchy mean value theorem to f,gf,g on the interval between aa and xx: f(x)−f(a)g(x)−g(a)=f′(ξ)g′(ξ)\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f'(\xi)}{g'(\xi)} for some ξ\xi strictly between aa and xx. As x→ax\to a, ξ→a\xi\to a too, so f(x)g(x)=f′(ξ)g′(ξ)→lim⁡t→af′(t)g′(t)\frac{f(x)}{g(x)}=\frac{f'(\xi)}{g'(\xi)}\to \lim_{t\to a}\frac{f'(t)}{g'(t)}. The ∞/∞\infty/\infty case follows from a similar but more delicate estimate.

Topics that use this theorem

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Step-by-step proofs

No step-by-step proof yet for this theorem.

References

  1. Guillaume François Antoine de L'Hôpital (1696). Analyse des infiniment petits, pour l'intelligence des lignes courbes
  2. Carl B. Boyer (1968). A History of Mathematics