A Kinder, Gentler, Longer L'Hospital Proof

[rudinreader]

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+ \in \mathbb{R}, 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 -> +\infty, and f'(x)/g'(x) -> L as x -> \infty, then f(x)/g(x) -> L.
proof:
Since g \neq 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 -> \infty, e_x -> 0, and f(x)/g(x) -> L.


Tricky case.
mini#3: If f,g -> \infty, and f'(x)/g'(x) -> L \in \mathbb{R} as x -> \infty, 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 = \infty
mini#4: If f,g -> \infty, and f'(x)/g'(x) -> \infty as x -> \infty then f/g -> \infty.
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 -> \infty.


***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- \in \mathbb{R}, 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 -> -\infty, and f'(x)/g'(x) -> L as x -> \infty, then f(x)/g(x) -> L.
proof: Again the left hand version of mini#2.

mini#7: If f,g -> \infty, and f'(x)/g'(x) -> L \in \mathbb{R} as x -> -\infty, then f/g -> L.
proof: Left hand limit of mini#3.

mini#8: If f,g -> \infty, and f'(x)/g'(x) -> \infty as x -> -\infty then f/g -> \infty.
proof: Left hand limit of mini#4.

mini#9: If f,g -> \infty, and f'(x)/g'(x) -> -\infty as x -> +/-\infty then f/g -> \infty.
proof: Left/Right hand limits of mini#4 except f'/g' -> -\infty 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.

 

[rudinreader]

I missed some... Any other omissions/mistakes, please point i out or post corrections..

The tricky case (#3) with c real
mini#10: If f,g -> \infty, and f'(x)/g'(x) -> L \in \mathbb{R} as x -> c+ \in \mathbb{R}, then f/g -> L.
proof:
(This is also just a direct switch of c < x <= M instead of x >= M in #3.)

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

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

Then whenever c < 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.

mini#11: The c- left hand limit as in #10 is done in the same way from the left side.

mini#12: mini#4 when x -> c+ \in \mathbb{R}.
proof:
Fix N > 0. Choose M > c close enough to c so that f'(x)/g'(x) > N + 1 when c < x < M. Then for c < x < M choose E_x in (x,M) such that (g(x)-g(M))f'(E_x) = (f(x)-f(M))g'(E_x). Then as x -> c+, 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 -> \infty.

mini#13: mini#12 when x -> c-. Same, just approach c from the left.

 

[rudinreader]

One point is that the cases x -> c+ and x -> c- can be merged together and proved as a two sided limit.. In these cases, as I wrote it above I only wrote out the proof for one side.

As long as I got them all up there, there are 4 distinct approaches (mini#1,#2,#3,#4), and the other cases are all done from the same idea as long as you can make the switch between c < x <= M instead of x >= M, and left/right hand limit etc.

I get the impression that the proof in baby Rudin is meant to proof LHospital via only two cases, except the idea involved is more difficult, whereas the proofs I gave above are all very straightforward. I felt that using a "difficult idea" to simplify it to two cases (instead of just one) is not worth the price in comparison to 4 easy ideas for 4 cases.

 

[rudinreader]

I should finish off by organizing around the ideas, a 4 part L'Hospital.

Let R* be the real numbers with -\infty,+\infty added. Assume f,g are differentiable except at c, and g,g' are "nonzero near c".

Thm 1,2: 0/0 cases

Thm 1: If f,g -> 0 as x -> c in R, and f'/g' -> L in R*, then f/g -> L.
proof: as in mini#1. The CMVT(Cauchy mean value theorem) gives f(x)/g(x) = f'(E_x)/g'(E_x), where E_x is squeezed between x and c.

Thm 2: If f,g -> 0, and f'/g' -> L in R* as x -> +/-\infty, then f/g -> L.
proof: as in mini#2. Choose M so that f'(x)/g'(x) is "near L" when x is between M and c. For each x choose M_x between x and c so that f(x)/g(x) ≈ (f(x)-f(M_x))/(g(x)-g(M_x)). The CMVT gives f(x)/g(x) ≈ f'(E_x)/g'(E_x) where E_x is squeezed between x and c.


Thm 3,4: infinity/infinity cases

Thm 3: If f,g -> \infty, and f'(x)/g'(x) -> L in R as x -> c in R*, then f/g -> L.
proof: as in mini#3. Use the algebraic identity so you can damp unwanted terms as g(x) -> infinity. Choose M "close" to c so that f'(M)/g'(M) ≈ L. Then choose M1 "closer" to c (than M) so that g(x) damps unwanted terms when x is between M1 and c. Apply CMVT with M, but with algebra we get f(x)/g(x) ≈ L when x is between M1 and c (because the unwanted terms are "damped by g(x)" there).

Thm 4: If f,g -> \infty, and f'(x)/g'(x) -> \infty as x -> c in R*, then f/g -> \infty.
proof: as in mini#4. Fix N > 0. Choose M "close enough to c" so that f'(x)/g'(x) > N when x is between M and c. Then as x -> c+, f(x)/g(x) ≈ (f(x)-f(M))/(g(x)-g(M)), so apply the CMVT and get f(x)/g(x) ≈ f'(E_x)/g'(E_x) where E_x is squeezed between x and c.

OK That's enough!