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Find the limit.

  1. Feb 5, 2008 #1
    Find the limit.

    limit x -> 0 cos(pi*x / sinx)

    I'm only in cal I. I was hoping I wouldn't have to use the difference quotient so is there an easier way to do this using derivative rules or something?
  2. jcsd
  3. Feb 5, 2008 #2


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    [tex]lim_{x\rightarrow 0} cosf(x) =cos (lim_{x\rightarrow 0} f(x))[/tex]
  4. Feb 5, 2008 #3
    [tex]\lim_{x\rightarrow 0}\cos{\left(\frac{\pi x}{\sin x}\right)}[/tex]

  5. Feb 5, 2008 #4
    Yeah this is what he meant, i guess. And follow rock.freak667's advice. But remember that this is true only because cos is continuous for any real.
    [tex]lim_{x\rightarrow 0} cosf(x) =cos (lim_{x\rightarrow 0} f(x))[/tex]
    Now can you see the trick after you go inside the cosine function with the limit?
  6. Feb 5, 2008 #5

    Can I just use the quotient rule on the inside?
  7. Feb 5, 2008 #6
    well you do not need to use the quotient rule, because do you know what the limit of

    sin(x)/x is as x--->0

    lim(x-->0)(sin(x))/x ----?????
    Just use this fact and you will be fine, because if you use the quotient rule as you are claimint to, you will get an intermediate form of 0/0.
  8. Feb 5, 2008 #7
    This would be x / sin x in this case. You got it backwards.
  9. Feb 5, 2008 #8
    NO, i did not get it backwards, because if you know what lim(x-->0)(sin(x))/x ----?????
    is you will not have problem finding out what
    lim(x-->0)x/sinx is, since you will take its reciprocal and everything will turn into terms of sin(x)/x
  10. Feb 5, 2008 #9
    Oh, sorry, calm down lol. It's 1.
  11. Feb 5, 2008 #10
    [tex]\lim_{x\rightarrow 0}\cos{\left(\frac{\pi x}{\sin x}\right)}=\cos{(\lim_{x\rightarrow 0}\left\frac{\pi}{\frac{\sin x}{\ x}\right)}}[/tex]
    do u see now what to do?
  12. Feb 5, 2008 #11
    Ya thanks it's -1.
  13. Feb 5, 2008 #12
    Good job!
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