Does lim┬(x→0)〖2xtanx/(1-e^x )^2〗and if it does, answer it.
The limit is 2,but not as demonstrated above.Since both the numerator and denominator tend to zero as x tends to zero,we need to use L'Hopital's theorem.
we have f(x)/g(x) =f'(x)/g'(x)
lim(x) tends to zero lim(x) tends to zero, and we can repeat until
we get rid of any zero's on the denominator.
Applying the theorem twice gets you there.