Let f(x)/g(x) be infinite, that is to say, x tends to a, and f(x) and g(x) tend to infinity.
Then the corresponding 0/0 type is
F (x)/g (x) = (1/g (x))/(1/f (x)), the right side of the equation is 00 type, and the right side is regular.
The formula = [g' (x)/g 2 (x)]/[f' (x)/f 2 (x)] is very clear. The problem of format is abstract.
After simplification, on the right side of the sum equation is = [g' (x)/f' (x)] * [f (x)/g (x)] 2.
Finally f(x)/g(x)=f'(x)/g'(x)
Obtain a certificate
It can also be proved by the mean value theorem, which is more troublesome.