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anonymous
 5 years ago
limit of (2x(csc4x)^2) as x approaches 0
anonymous
 5 years ago
limit of (2x(csc4x)^2) as x approaches 0

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02x is zero. Brush up on a lil trig what is csc 0?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 2x is zero wouldn't it all be zero?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In a finite world, yes, but you are currently residing in a potentially infinite world: possibility of 0 times infinity=indeterminate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when using l'hospital's rule don't we take the derivative of each or is that wrong?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't know that you need. It seems the lim is 0 times 1; zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh no, sin is 0, yeah lhopital

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think it is chain rule. I think it is only when one on top, one on bottom, you do it the other way.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah i'm thinking this specific problem is chain rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, that makes sense. Thank you!
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