Famous Theorems of Mathematics/L'Hôpital's rule

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Consider some limit

 \lim_{x\to a} \frac{f(x)}{g(x)}

which cannot be evaluated directly, that is, either f(a) = g(a) = 0 or

 \lim_{x\to a} g(x) = \lim_{x\to a} g(x) =  \infty

L'Hopital's rule says that

 \lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)}

Suppose both are equal to zero. The Newton definition of the derivative is

 f'(a) = \lim_{x\to a} \frac{f(x) - f(a)}{x-a}

Therefore,

 \frac{f'(a)}{g'(a)} =  \lim_{x\to a} \frac{\frac{f(x) - f(a)}{x-a}}{\frac{g(x) - g(a)}{x-a}} = \lim_{x\to a} \frac{f(x) - f(a)}{g(x) - g(a)}

 \frac{f'(a)}{g'(a)} = \frac{f(a)}{g(a)} = \lim_{x\to a} \frac{f'(a)}{g'(a)}

Now suppose that both  f and  g diverge to positive or negative infinity. Another way to define the derivative is by

 f'(a) = \lim_{x\to 0} \frac{f(x+a)}{x}

Then

 \frac{f'(a)}{g'(a)}  = \lim_{x\to 0} \frac{xf(x+a)}{xg(x+a)} = \lim_{x\to 0} \frac{f(x+a)}{g(x+a)} = \frac{f(x+a)}{g(x+a)}

QED