1. ## Limits with series

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?

2. We could expand it and then apply L'Hopital to each term.

$\lim_{x\to 0}\frac{sin^{2}(3x)}{(e^{2x}-2x-1)^{3}}-6\lim_{x\to 0}\frac{xsin(3x)}{(e^{2x}-2x-1)^{3}}+9\lim_{x\to 0}\frac{x^{2}}{(e^{2x}-2x-1)^{3}}$

If L'Hopital is applied 6 times to the left one, we get 81/20.

The middle one gives 243/160

The last one gives 0.

$\frac{81}{20}-\frac{243}{160}=\frac{81}{32}$

It is rather cumbersome. I am sure there is a more clever way to go about it.

3. Ah yes, we're not allowed to use L'H Rule at all though. Really I think my error is just in finding the Maclaurin expansion of the numerator...can someone just check if they can spot what I am doing wrong there though?

4. CANCEL This problem...I got the answer.