1-a) Suppose f, g are differentiable in (a,infinity), g'(x) != 0 on (a,infinity),
f'(x)/g'(x) -> L in R as x -> infinity, and g(x) -> infinity as x -> infinity.
Then f(x)/g(x) -> L as x -> infinity.
proof:
The trick is to use 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. Let x_n be a sequence in (a,infinity) such that x_n -> infinity. Fix N0 large enough so that f'(x_n)/g'(x_n) is in (L-e,L+e) for n >= N0. Then fix N1 >= N0 large enough so that x_n > x_N0 for n >= N1, and both f(x_N0)/g(x_n), g(x_N0)/g(x_n) are in (-e,e) for n >= N1.
For each n >= N1, choose E_n in (x_N0,x_n) such that
f'(E_n)/g'(E_n) = (f(x_n)-f(x_N0))/(g(x_n)-g(x_N0)).
Then for all n >= N1,
|f(x_n)/g(x_n) - L|
= |f(x_N0)/g(x_n) + (1-g(x_N0)/g(x_n))*f'(E_n)/g'(E_n) - L|
<= |f(x_N0)/g(x_n)| + |f'(E_n)/g'(E_n) - L| + |g(x_N0)/g(x_n)*f'(E_n)/g'(E_n)|
< e + e + |g(x_N0)/g(x_n)|*(|L| + e)
< 2e + e(|L| + e) = e(|L| + 2) + e^2.
The right hand side can be made arbitrarily small, so we conclude f(x_n)/g(x_n) -> L as n -> infinity. Since {x_n} was chosen arbitrarily, we are done.
1-b) Suppose f, g are differentiable in (a,infinity), g'(x) != 0 on (a,infinity),
f'(x)/g'(x) -> infinity as x -> infinity, and g(x) -> infinity as x -> infinity.
Then f(x)/g(x) -> infinity as x -> infinity.
proof:
Again, use the trick
f(x)/g(x) = f(y)/g(x) + (1-g(y)/g(x))*(f(x)-f(y))/(g(x)-g(y)).
Fix M > 0. Let x_n be a sequence in (a,infinity) such that x_n -> infinity. Fix N0 large enough so that f'(x_n)/g'(x_n) > M for n >= N0. Then fix N1 >= N0 large enough so that x_n > x_N0 for n >= N1, and both f(x_N0)/g(x_n), g(x_N0)/g(x_n) are in (-e,e) for n >= N1.
For each n >= N1, choose E_n in (x_N0,x_n) such that
f'(E_n)/g'(E_n) = (f(x_n)-f(x_N0))/(g(x_n)-g(x_N0)).
Then for all n >= N1,
f(x_n)/g(x_n)
= f(x_N0)/g(x_n) + (1-g(x_N0)/g(x_n))*f'(E_n)/g'(E_n)
(use small e < 1)
> -e + (1 - e)M.
The right hand side can be made arbitrarily large, so we conclude f(x_n)/g(x_n) -> infinity as n -> infinity. Since {x_n} was chosen arbitrarily, we are done.
2-a) Suppose f, g are differentiable in (a,b), g'(x) != 0 on (a,b), f(x) -> 0, g(x) -> 0 as x -> a, and f'(x)/g'(x) -> L in R as x -> a. Then f(x)/g(x) -> L as x -> a.
Proof:
First expand f, g continuously (or redefine them) so that f(a) = g(a) = 0. Fix e > 0. Let x_n be a sequence in (a,b) such that x_n -> a. For each n >= N0, choose E_n in (a,x_n) such that f'(E_n)/g'(E_n) = (f(x_n)-f(a))/(g(x_n)-g(a)) = f(x_n)/g(x_n). Clearly E_n -> a. Thus f(x_n)/g(x_n) = f'(E_n)/g'(E_n) -> L as n -> infinity.