- #1

rudinreader

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I have found the easiest way to understand the proof is to give attention to a lot of particular cases. So in that since, L'Hospital's rule is a compact theorem statement summarizing a handful of situations. That can make it's proof (which part?) hard to remember. Baby Rudin makes it hard to see even the easiest case (below). The "proof" on wiki is just a couple of lines, and isn't really complete. Perhaps this can be posted there... hmm... anyways...

Theorem: L'Hospital's Rule

To see the theorem statement, go to wiki, "Formal Statement": http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule.

proof(s) (of each situation):

With the statement "as x -> c", in each case, assume g(x), g'(x) are "nonzero near c", as discussed on the wiki page. f,g are assumed differentiable besides at c.

The easiest case first.

mini#1:If f(x),g(x) -> 0 as x -> c+ [itex]\in \mathbb{R}[/itex], and f'(x)/g'(x) -> L (possibly infinite) as x -> c+, then f(x)/g(x) -> L.

proof:

Assume f,g are continuous at c (or redefine them as f(c) = 0, etc). If x > c, then by the CMVT, there exists E_x in (c,x) such that f(x)g'(E_x) = g(x)f'(E_x), so that as x -> c+, f(x)/g(x) = f'(E_x)/g'(E_x) -> L.

The easiest case modified.

mini#2:If f(x),g(x) -> 0 as x -> +[itex]\infty[/itex], and f'(x)/g'(x) -> L as x -> [itex]\infty[/itex], then f(x)/g(x) -> L.

proof:

Since g [itex]\neq[/itex] 0, for each x we can choose M_x > x large enough such that (f(x)-f(M_x))/(g(x)-g(M_x)) is as close as we want to f(x)/g(x), say within e_x = 1/x > 0. Then we can choose E_x in (x,M_x) such that (f(E_x)-f(x))g'(E_x) = (g(M_x)-g(x))f'(E_x), so that |f(x)/g(x) - f'(E_x)/g'(E_x)| < e_x, so when x -> [itex]\infty[/itex], e_x -> 0, and f(x)/g(x) -> L.

Tricky case.

mini#3:If f,g -> [itex]\infty[/itex], and f'(x)/g'(x) -> L [itex]\in \mathbb{R}[/itex] as x -> [itex]\infty[/itex], then f/g -> L.

proof:

Verify the equality f(x)/g(x) = f(y)/g(x) + (1-g(y)/g(x))*(f(x)-f(y))/(g(x)-g(y)).

Fix e > 0. Fix M > 0 such that x >= M implies |f'(x)/g'(x)-L| < e. Since f(M), g(M) are fixed, we can choose M1 > M such that x >= M1 implies |f(M)/g(x)|,|g(M)/g(x)| < e.

With this set up apply the CMVT, for any x > M, choose Cx in (M,x) such that f'(Cx)/g'(Cx) = (f(x)-f(M))/(g(x)-g(M)).

Then for all x >= M1,

|f(x)/g(x) - L|

= |f(M)/g(x) + (1-g(M)/g(x))*f'(Cx)/g'(Cx) - L|

<= |f(M)/g(x)| + |f'(Cx)/g'(Cx) - L| + |g(M)/g(x)*f'(Cx)/g'(Cx)|

< e + e + |g(M)/g(x)|*(|L| + e)

< 2e + e(|L| + e) = e(|L| + 2) + e^2.

It follows that f(x)/g(x) -> L.

Another trick for L = [itex]\infty[/itex]

mini#4:If f,g -> [itex]\infty[/itex], and f'(x)/g'(x) -> [itex]\infty[/itex] as x -> [itex]\infty[/itex] then f/g -> [itex]\infty[/itex].

proof:

Fix N > 0. Choose M large enough so that f'(x)/g'(x) > N + 1 for x > M. Then for x > M choose E_x such that (g(x)-g(M))f'(E_x) = (f(x)-f(M))g'(E_x). Then as x gets large, f(x)/g(x) ≈ (f(x)-f(M))/(g(x)-g(M)) = f(E_x)/g(E_x) > N. Since N was arbitrary, we conclude f/g -> [itex]\infty[/itex].

***Extra justification for the statement (for beginners..): "as x gets large, f(x)/g(x) ≈ (f(x)-f(M))/(g(x)-g(M))".

Fix e > 0. Since f,g -> infinity, choose e*f(x), e*g(x) larger than f(M), g(M), then ..

(1+e)f(x) > f(x) - f(M) > (1-e)f(x), etc., so (f(x)/g(x))*((1-e)/(1+e)) < f(x)-f((M_x))/(g(x)-g(M_x)) < (f(x)/g(x))*((1+e)/(1-e)) -> f(x)/g(x) as e -> 0.

Other cases

mini#5:If f(x),g(x) -> 0 as x -> c- [itex]\in \mathbb{R}[/itex], and f'(x)/g'(x) -> L (possibly infinite) as x -> c+, then f(x)/g(x) -> L.

proof: This is the left hand limit of mini#1, the same argument applies (only it's on the left hand side).

mini#6:If f(x),g(x) -> 0 as x -> -[itex]\infty[/itex], and f'(x)/g'(x) -> L as x -> [itex]\infty[/itex], then f(x)/g(x) -> L.

proof: Again the left hand version of mini#2.

mini#7:If f,g -> [itex]\infty[/itex], and f'(x)/g'(x) -> L [itex]\in \mathbb{R}[/itex] as x -> -[itex]\infty[/itex], then f/g -> L.

proof: Left hand limit of mini#3.

mini#8:If f,g -> [itex]\infty[/itex], and f'(x)/g'(x) -> [itex]\infty[/itex] as x -> -[itex]\infty[/itex] then f/g -> [itex]\infty[/itex].

proof: Left hand limit of mini#4.

mini#9:If f,g -> [itex]\infty[/itex], and f'(x)/g'(x) -> -[itex]\infty[/itex] as x -> +/-[itex]\infty[/itex] then f/g -> [itex]\infty[/itex].

proof: Left/Right hand limits of mini#4 except f'/g' -> -[itex]\infty[/itex] this time. The same argument in #4 applies except you fix N < 0, and show f(x)/g(x) < N for large (or large negative) x.

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