L'Hopital's Rule Lim Help

moloko

1. Homework Statement
lim as x->0 (tan(x)-x)/(sin2x-2x)

2. Homework 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!

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Dick

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

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

Dick

Homework Helper
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.

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!

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