We need to compute the limit as x-> 0 of

(sin 3x -3x)^2/(e^2x -1 - 2x) ^3

Now, according to Wolfram, this is 81/32

I used the Maclaurin series for e^x and took the first term after applying the -1-2x and got (4x^2)^3x^6/2! and got 32 x^6

Now that will give the correct denominator.

For the numerator I used the second term in the Maclaurin series of sin(3x) and squared it. This gave (-27)^2*x^6/3!, but this 121.5 x^6, but it should give 81 so that the limit is 81/32. What am doing wrong in this step? How do you get the 81?

The it asks how many times L'Hopital's Rule would have to be used to compute this? Is the answer just 6 since we have the term x^6 present?