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.