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The standard statement of l'Hôpital's rule requires that the limit as x -> a of f'/g' exists, which in particular means that f' and g' must exist in a neighbourhood of a, and Cauchy's mean value theorem can be used to prove it.
Here is a restatement, for the 0/0 case only, with proof involving only the definition of derivative, that only requires that f' and g' exist, and that g' is non-zero, at the point a, followed by a generalisation that requires that the limits of f(m) and g(m) exist for m < n, and that f(n)(a) and g(n)(a) exist, and that g(n)(a) is non-zero:
l'Hôpital's theorem, with all its conditions, and the use of Cauchy's mean value theorem, is still required for the ∞/∞ case, or for infinite a
tîny-tîm's theorem:
For any functions f and g with [itex]\lim_{x\to a}f(a) = \lim_{x\to a}g(a) = 0[/itex] for which f'(a) and g'(a) exist with g'(a) ≠ 0:
[tex]\lim_{x\to a}\frac{f(x)}{g(x)}\ =\ \lim_{x\to a}\frac{f(x)}{x-a}\frac{x-a}{g(x)}\ =\ \lim_{x\to a}\frac{f(x)}{x-a}\lim_{x\to a}\frac{x-a}{g(x)}\ =\ \frac{f'(a)}{g'(a)}[/tex]
(because the limit of the quotient is the quotient of the limits, provided that the denominator limit is non-zero)
generalisation:
For any functions f and g with [itex]\lim_{x\to a}f^{(m)}(a) = \lim_{x\to a}g^{(m)}(a) = 0\ \forall m : 0 < m < n[/itex] for which f(n)(a) and g(n)(a) exist with g(n)(a) ≠ 0:
[tex]\lim_{x\to a}\frac{f(x)}{g(x)}\ =\ \lim_{x\to a}\frac{(f(x))^n}{(x-a)^n}\frac{(x-a)^n}{(g(x))^n}\ =\ \frac{f^{(n)}(a)}{g^{(n)}(a)}[/tex]
corollary:
If p(0) = q(0) = f(a) = g(a) = 0, and if f'(a) and g'(a) exist and are non-zero, and if the final limit below exists, then:
(by assumption, there is a neighbourhood of a in which f and g are non-zero, so: )
[tex]\lim_{x\to a} \frac{p(f(x))}{q(g(x))}\ =\ \lim_{x\to a}\frac{p(f(x))}{f(x)}\lim_{x\to a}\frac{f(x)}{x\ -\ a}\lim_{x\to a}\frac {x\ -\ a}{g(x)}\lim_{x\to a}\frac{g(x)}{q(g(x))} \ =\ \frac{f'(a)}{g'(a)} \lim_{x\to a} \frac{p'(f(x))}{q'(g(x))}[/tex]
and generally, if p(0) = q(0) = 0, and if [itex]\lim_{x\to a}f^{(m)}(a) = \lim_{x\to a}g^{(m)}(a) = 0\ \forall m : 0 < m < n[/itex], and if f(n)(a) and g(n)(a) exist and are non-zero, and if the final limit below exists, then:
[tex]\lim_{x\to a} \frac{p(f(x))}{q(g(x))}\ =\ \lim_{x\to a}\frac{p(f(x))}{f(x)^n}\left(\lim_{x\to a}\frac{f(x)}{x\ -\ a}\right)^n\left(\lim_{x\to a}\frac {x\ -\ a}{g(x)}\right)^n [/tex]
Here is a restatement, for the 0/0 case only, with proof involving only the definition of derivative, that only requires that f' and g' exist, and that g' is non-zero, at the point a, followed by a generalisation that requires that the limits of f(m) and g(m) exist for m < n, and that f(n)(a) and g(n)(a) exist, and that g(n)(a) is non-zero:
l'Hôpital's theorem, with all its conditions, and the use of Cauchy's mean value theorem, is still required for the ∞/∞ case, or for infinite a
tîny-tîm's theorem:
For any functions f and g with [itex]\lim_{x\to a}f(a) = \lim_{x\to a}g(a) = 0[/itex] for which f'(a) and g'(a) exist with g'(a) ≠ 0:
[tex]\lim_{x\to a}\frac{f(x)}{g(x)}\ =\ \lim_{x\to a}\frac{f(x)}{x-a}\frac{x-a}{g(x)}\ =\ \lim_{x\to a}\frac{f(x)}{x-a}\lim_{x\to a}\frac{x-a}{g(x)}\ =\ \frac{f'(a)}{g'(a)}[/tex]
(because the limit of the quotient is the quotient of the limits, provided that the denominator limit is non-zero)
generalisation:
For any functions f and g with [itex]\lim_{x\to a}f^{(m)}(a) = \lim_{x\to a}g^{(m)}(a) = 0\ \forall m : 0 < m < n[/itex] for which f(n)(a) and g(n)(a) exist with g(n)(a) ≠ 0:
[tex]\lim_{x\to a}\frac{f(x)}{g(x)}\ =\ \lim_{x\to a}\frac{(f(x))^n}{(x-a)^n}\frac{(x-a)^n}{(g(x))^n}\ =\ \frac{f^{(n)}(a)}{g^{(n)}(a)}[/tex]
corollary:
If p(0) = q(0) = f(a) = g(a) = 0, and if f'(a) and g'(a) exist and are non-zero, and if the final limit below exists, then:
(by assumption, there is a neighbourhood of a in which f and g are non-zero, so: )
[tex]\lim_{x\to a} \frac{p(f(x))}{q(g(x))}\ =\ \lim_{x\to a}\frac{p(f(x))}{f(x)}\lim_{x\to a}\frac{f(x)}{x\ -\ a}\lim_{x\to a}\frac {x\ -\ a}{g(x)}\lim_{x\to a}\frac{g(x)}{q(g(x))} \ =\ \frac{f'(a)}{g'(a)} \lim_{x\to a} \frac{p'(f(x))}{q'(g(x))}[/tex]
and generally, if p(0) = q(0) = 0, and if [itex]\lim_{x\to a}f^{(m)}(a) = \lim_{x\to a}g^{(m)}(a) = 0\ \forall m : 0 < m < n[/itex], and if f(n)(a) and g(n)(a) exist and are non-zero, and if the final limit below exists, then:
[tex]\lim_{x\to a} \frac{p(f(x))}{q(g(x))}\ =\ \lim_{x\to a}\frac{p(f(x))}{f(x)^n}\left(\lim_{x\to a}\frac{f(x)}{x\ -\ a}\right)^n\left(\lim_{x\to a}\frac {x\ -\ a}{g(x)}\right)^n [/tex]
[tex]\lim_{x\to a}\frac{(g(x))^n}{q(g(x))}\ =\ \left(\frac{f'(a)}{g'(a)}\right)^n \lim_{x\to a} \frac{p^{(n)}(f(x))}{q^{(n)}(g(x))}[/tex]
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