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In a finite world, yes, but you are currently residing in a potentially infinite world: possibility of 0 times infinity=indeterminate.
I think the answer is undefined not indeterminate, there is a difference. If you try to solve using l'hospital's rule, you end up getting an answer of 2/0 when plugging in 0. 2/0 is undefined not indeterminate. For example, one indeterminate form is 0/0.
It may not get to that point. According to her finding 1/sin (0) is 1. Indeterminate was just mentioned as a potential, the need to investigate csc 0.
when using l'hospital's rule don't we take the derivative of each or is that wrong?
I don't know that you need. It seems the lim is 0 times 1; zero
Oh no, sin is 0, yeah lhopital
I think it is chain rule. I think it is only when one on top, one on bottom, you do it the other way.
Yeah i'm thinking this specific problem is chain rule
You can use l'hospital when you come across indeterminate forms. For example, in this problem plugging in 0 gives us 2(0)/[sin(4(0))]^2 which is equal to 0/0, which is an indeterminate form. Consequently, we can use l'hospital to investigate whether or not the limit is solvable or undefined. We do this by taking the derivative of the top and the derivative of the bottom, which yields: 2/8sin(4x)cos(4x). If we now plug in o we get 2/8sin(4(0))cos(4(0)), which equals 2/0. 2/0 is undefinded and we cannot perform l'hospitals rule anymore; the answer is just undefined..i think anyway.