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anonymous

  • 4 years ago

@jamesJ or whoever knws, limit x->0 (2sinx-1)?? dont knw where to begin on this one..

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  1. asnaseer
    • 4 years ago
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    use the same method as James showed you in your last question.

  2. anonymous
    • 4 years ago
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    but what is sin? sin=1/cscx??

  3. myininaya
    • 4 years ago
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    f(x)=2 sin(x)-1 is continuous at x=0 therefore you just evaluate f(0) to find the limit here

  4. anonymous
    • 4 years ago
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    how can you tell it is continuous?

  5. myininaya
    • 4 years ago
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    because y=sin(x) is a continuous function It exists for any real input

  6. anonymous
    • 4 years ago
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    ok so on that note, is y=cosx also a continuous function?

  7. myininaya
    • 4 years ago
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    yep

  8. myininaya
    • 4 years ago
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    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

  9. myininaya
    • 4 years ago
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    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

  10. myininaya
    • 4 years ago
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    these functions* as in y=tan(x),sec(x),cot(x),csc(x)

  11. anonymous
    • 4 years ago
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    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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