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anonymous
 4 years ago
@jamesJ or whoever knws, limit x>0 (2sinx1)?? dont knw where to begin on this one..
anonymous
 4 years ago
@jamesJ or whoever knws, limit x>0 (2sinx1)?? dont knw where to begin on this one..

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asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0use the same method as James showed you in your last question.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but what is sin? sin=1/cscx??

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2f(x)=2 sin(x)1 is continuous at x=0 therefore you just evaluate f(0) to find the limit here

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how can you tell it is continuous?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2because y=sin(x) is a continuous function It exists for any real input

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so on that note, is y=cosx also a continuous function?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2but y=tan(x) and y=sec(x) and y=cot(x) and y=csc(x) have discontinuities because at some point there bottom is zero

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2like i'm saying there functions i just mentioned can be written in terms of cos(x) and sin(x) and remember a function is discontinuous( or one reason it can be discontinuous ) is if the bottom is zero

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2these functions* as in y=tan(x),sec(x),cot(x),csc(x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u r a mind reader haha, that was exactly the question i was going to ask. thank u for ur help!!!!!
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