Prove the following: L\'Hopital\'s Rule. f(x)/g(x) is close to (f(x) - f(t)/(g(x

Question

Prove the following:

L'Hopital's Rule. f(x)/g(x) is close to (f(x) - f(t)/(g(x) - g(t)) if t is really close to b which in turn is equal to f'(theta)/g'(theta) for some theta between x and t.

Explanation / Answer

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) Therefore whether we have 0/0 or infinity/infinity we can use l'Hopital's rule.

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