Use L'hospital's Rule to evaluate

Lim x -->0 9x(cos5x-1)/sin(2x-2x)

I know that this is a 0/0 indeterminate form and you need to differentiable both top and bottom until you can substitute 0 into x. I have problems differentiating sin(2x-2x).

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- Nov 15th 2009, 11:52 AMBelowzero78L'Hospital's Rule
Use L'hospital's Rule to evaluate

Lim x -->0 9x(cos5x-1)/sin(2x-2x)

I know that this is a 0/0 indeterminate form and you need to differentiable both top and bottom until you can substitute 0 into x. I have problems differentiating sin(2x-2x). - Nov 15th 2009, 12:31 PMScott H
Do you mean $\displaystyle \sin 2x - 2x$? If so, we may differentiate it as follows:

$\displaystyle \frac{d}{dx}(\sin 2x - 2x)=\frac{d}{dx}(\sin 2x)-\frac{d}{dx}(2x)=2\cos 2x-2.$ - Nov 15th 2009, 12:39 PMBelowzero78
Yes, thats write, so how to do compute the limit afterwards? i got something like -45x*sin5x+9cos5x-9/2cos2x-2

This would still result in [0,0]. So i would differentiate once again. - Nov 15th 2009, 12:57 PMBelowzero78
my final answer i got is 675/8, I want to check if this is correct. Thanks to everyone!