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To prove L'Hopital's Rule (sometimes spelled L'Hospital's Rule), we
use the Taylor expansion:

f(a+h) = f(a) + hf'(a) + terms in h^2 and higher
g(a+h) = g(a) + hg'(a) + terms in h^2 and higher

So:

f(a+h) f(a)+h*f'(a)
Lt ------ -> ------------
h->0 g(a+h) g(a)+h*g'(a)

so with f(a) = g(a) = 0 we get:

f(a+h) h*f'(a) f'(a)
Lt ------- -> ------- -> ------
h->0 g(a+h) h*g'(a) g'(a)

We can use l'Hopital's also if f'(a) -> infinity and g'(a) -> infinity:

f(a) infinity 1/g(a) 0
---- -> -------- so -------- -> ---
g(a) infinity 1/f(a) 0

and applying l'Hopital's to this latter expression, we get:

f(a) -g'(a)/[g(a)]^2 g'(a)*[f(a)]^2
------ -> ---------------- -> ----------------
g(a) -f'(a)/[f(a)]^2 f'(a)*[g(a)]^2

and cross-multiplying:

f'(a) f(a)
------- -> ------
g'(a) g(a)

why dont you just divide both num and denominator by x^1/2? then its done without L hospital ..