lim┬(x→0)⁡〖2xtanx/(1-e^x )^2〗?

The limit is 2:

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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.                    

Thanks Mathcad

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