anonymous
  • anonymous
@jamesJ or whoever knws, limit x->0 (2sinx-1)?? dont knw where to begin on this one..
Mathematics
chestercat
  • chestercat
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asnaseer
  • asnaseer
use the same method as James showed you in your last question.
anonymous
  • anonymous
but what is sin? sin=1/cscx??
myininaya
  • myininaya
f(x)=2 sin(x)-1 is continuous at x=0 therefore you just evaluate f(0) to find the limit here

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anonymous
  • anonymous
how can you tell it is continuous?
myininaya
  • myininaya
because y=sin(x) is a continuous function It exists for any real input
anonymous
  • anonymous
ok so on that note, is y=cosx also a continuous function?
myininaya
  • myininaya
yep
myininaya
  • myininaya
but 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
  • myininaya
like 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
  • myininaya
these functions* as in y=tan(x),sec(x),cot(x),csc(x)
anonymous
  • anonymous
u r a mind reader haha, that was exactly the question i was going to ask. thank u for ur help!!!!!

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