# L'Hopital's Rule Help

1. Apr 4, 2007

### moloko

1. The problem statement, all variables and given/known data
lim as x->0 (tan(x)-x)/(sin2x-2x)

2. Relevant equations
L'Hopitals rule states that if the limit reaches 0/0, you can take the derivative of the top and the bottom until you get the real limit.

3. The attempt at a solution

(sec^2(x)-1)/(2cos2x-2) still 0/0
2sec^2(x)tan(x)/(-4sin(2x)) still 0/0

I have pain stakingly taken the derivative twice more and it simply does not seem to reach any end. All help is very much appreciated!

2. Apr 4, 2007

### Dick

You are going to get a nonzero denominator at the next derivative after you've shown. It's a cosine.

3. Apr 4, 2007

### moloko

You are correct and this does give me a non 0/0 answer but the answer is wrong according to the book. It should be -1/4.

My next derivative is:

(2*(2sec^2(x)tan(x))*tan(x))/(-8cos(2x))

In simpler terms

(4sec^2(x)tan^2(x))/(-8cos(2x))

The numerator becomes 0 with the limit and the demoninator becomes -8, making the limit 0.

Thanks

4. Apr 4, 2007

### Dick

Why aren't you using the product rule on the numerator? What ARE you doing? The derivative of the tan(x) term will be nonzero.

5. Apr 4, 2007

### moloko

I apologize, that was a careless mistake.

I'll post my solution if it's of any help to anyone...

the numerator becomes 2(sec^2(x)tan^2(x) + sec^2(x)),
which of course goes to 2 when pushed to 0
leaving 2/-8, or -1/4

Thank you so much!