If f and g are differentiable in the open interval containing L [which may be a finite limit or ±∞] and if
limitx→L f'(x)/g'(x) exists,
then the indeterminate form:
limitx→L f(x)/g(x)
where f and g are both zero, or f and g are both ±∞ also exist, and
limitx→L f(x)/g(x) = limitx→L f'(x)/g'(x)
So, just for example: with f(x) = x2 g(x) = 3x2. Both are differentiable, limitx→0 f(x)/g(x) = x2/3x2 which → 0/0, an indeterminate form.
Differentiate twice: limitx→0x2/3x2 = limitx→0 2x/6x = limitx→0 2/6 = 1/3.
Obviously, you could get this answer just by "factoring out" x2. Just algebra; no Calculus required.
However, you can't factor this one: limitx→0 sin(x)/x.
L'Hospital's Rule gives:
limitx→0 sin(x)/x = limitx→0 cos(x)/1 = 1.
Remember to apply L'Hospitals Rule: you don't do the rule for differentiating a quotient.. That would give [f'(x)g(x) - g'(x)f(x)]/[g'(x)]2. You simply take f'(x)/g'(x) and check the limit.
As long as f, f', f'' and g, g', g'' [etc] are still differentiable and their quotient is indeterminate, you can apply the rule as many times as necessary to get an answer.