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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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).
Do you mean? If so, we may differentiate it as follows:
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.
my final answer i got is 675/8, I want to check if this is correct. Thanks to everyone!