L’Hôpital’s Rule
Let a,L∈R∪{−∞,∞} be extended real numbers, and let I be an open interval containing a(if a∈R) or having a as an endpoint(if a=±∞)
Let f,g:I∖{a}→R be differential functions on I∖{a} such that
g′(x)=0for allx∈I∖{a}
Suppose that one of the following conditions holds:
- limx→af(x)=0andlimx→ag(x)=0
- limx→a∣g(x)∣=∞
If the limit of the raio of their derivatives exists, that is,
x→alimg′(x)f′(x)=L
Then the limit of ratio of the functions exists and is given by:
x→alimg(x)f(x)=L
Proof of the 0/0 case
First suppose that a,L∈R
we may define f(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⟹g′(y)f′(y)−L<ε
Take x such that
0<∣x−a∣<δ.
By the Cauchy Mean Value Theorem, there exists c strictly between a and x such that
g(x)−g(a)f(x)−f(a)=g′(c)f′(c).
Since f(a)=g(a)=0,
g(x)f(x)=g′(c)f′(c).
Since c lies between a and x,
0<∣c−a∣<∣x−a∣<δ.
Therefore,
g(x)f(x)−L=g′(c)f′(c)−L<ε.
Hence,
x→alimg(x)f(x)=L.
Proof of other case
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