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Calculus
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L’Hôpital’s Rule

Let a,L∈R∪{−∞,∞}a,L\in\mathbb{R}\cup\{-\infty,\infty\} be extended real numbers, and let II be an open interval containing a (if a∈R)a\,(\text{if }a\in\mathbb{R}) or having aa as an endpoint (if a=±∞)\,(\text{if }a=\pm\infty)

Let f,g:I∖{a}→Rf,g:I\setminus\{a\}\to\mathbb{R} be differential functions on I∖{a}I\setminus\{a\} such that g′(x)≠0for allx∈I∖{a}g'(x)\neq 0 \quad \text{for all} \quad x\in I\setminus\{a\}

Suppose that one of the following conditions holds:

  1. lim⁡x→af(x)=0andlim⁡x→ag(x)=0\lim_{x \to a} f(x) = 0 \quad \text{and} \quad \lim_{x \to a} g(x) = 0
  2. lim⁡x→a∣g(x)∣=∞\lim_{x \to a} \lvert g(x) \rvert= \infty

If the limit of the raio of their derivatives exists, that is,

lim⁡x→af′(x)g′(x)=L\lim_{x \to a}\frac{f'(x)}{g'(x)}=L

Then the limit of ratio of the functions exists and is given by:

lim⁡x→af(x)g(x)=L\lim_{x \to a}\frac{f(x)}{g(x)}=L

Proof of the 0/00/0 case

First suppose that a,L∈Ra,L\in\mathbb{R}

we may define f(a)=g(a)=0f(a)=g(a)=0 with these definitions, f and g are continuous at a

By definitions of a limit,

∀ε>0, ∃δ>0s.t.0<∣y−a∣<δ⟹∣f′(y)g′(y)−L∣<ε\forall \varepsilon>0,\ \exists \delta>0 \quad \text{s.t.} \quad 0<\lvert y-a\rvert<\delta \quad \Longrightarrow \quad \left|\frac{f'(y)}{g'(y)}-L\right|<\varepsilon

Take xx such that

0<∣x−a∣<δ.0<\lvert x-a\rvert<\delta.

By the Cauchy Mean Value Theorem, there exists cc strictly between aa and xx such that

f(x)−f(a)g(x)−g(a)=f′(c)g′(c).\frac{f(x)-f(a)}{g(x)-g(a)} = \frac{f'(c)}{g'(c)}.

Since f(a)=g(a)=0f(a)=g(a)=0,

f(x)g(x)=f′(c)g′(c).\frac{f(x)}{g(x)} = \frac{f'(c)}{g'(c)}.

Since cc lies between aa and xx,

0<∣c−a∣<∣x−a∣<δ.0<\lvert c-a\rvert<\lvert x-a\rvert<\delta.

Therefore,

∣f(x)g(x)−L∣=∣f′(c)g′(c)−L∣<ε.\left|\frac{f(x)}{g(x)}-L\right| = \left|\frac{f'(c)}{g'(c)}-L\right| <\varepsilon.

Hence,

lim⁡x→af(x)g(x)=L.\lim_{x\to a}\frac{f(x)}{g(x)}=L.

Proof of other case

Please search it on internet


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