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Homework Help: Limit problem of death

  1. Dec 9, 2007 #1
    [Solved] Limit problem of death

    [tex] \lim_{x\rightarrow{\pi/2}} (x*tan(x) - \frac{\pi}{2*cos(x)}) [/tex]

    OK. I'm not sure how to begin this problem because:

    1) tan(pi/2) is undefined and

    2) pi/2cos(pi/2) is undefined also!

    Any hint would be great!

    Thanks
     
    Last edited: Dec 9, 2007
  2. jcsd
  3. Dec 9, 2007 #2

    Dick

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    Combine them into a single fraction and apply l'Hopital.
     
  4. Dec 9, 2007 #3
    like this?

    [tex] \frac{2cos(x)*x*tan(x)-\pi}{2cos(x)} [/tex]

    This looks incorrect...
     
  5. Dec 9, 2007 #4

    cristo

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    You can write tanx=sinx/cosx, so try to do that with your expression in the first post.
     
  6. Dec 9, 2007 #5

    Dick

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    Or just put it into your current version. cos(x)*tan(x)=sin(x).
     
  7. Dec 9, 2007 #6

    cristo

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    :rofl: yup, that works as well, and is a lot simpler to do!
     
  8. Dec 9, 2007 #7
    Ok, I'm still confused. I must be overseeing something obvious.

    I get

    [tex] \frac{xsin(x)- \pi}{cos{x}} [/tex]

    That doesn't work for l'Hôpital so I must be doing something wrong.
     
  9. Dec 9, 2007 #8

    Dick

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    What happened to the '2's? It should be a 0/0 limit.
     
  10. Dec 9, 2007 #9
    This is what I did

    [tex] \frac{2cos(x)*x*\frac{sinx}{cosx}-\pi}{2cos(x)} [/tex]

    [tex] \frac{2cos(x)*x*sin(x)-\pi*cos(x)}{cos(x)} * \frac{1}{2cos(x)} [/tex]

    [tex] \frac{2cos(x)*x*sin(x)-\pi*cos(x)}{2cos^2(x)} [/tex]

    Another result and they are both 0/0 but is this correct ?
     
  11. Dec 9, 2007 #10
    Ok I think I got it! The above equals to

    [tex] \frac{x*sin(x)-\pi/2}{cos(x)} : [\frac{0}{0}][/tex]
     
  12. Dec 9, 2007 #11
    Ok thanks a bunch guys. The rest is simple

    [tex] \lim_{x\rightarrow\pi/2} \frac{sin(x)+x*cos(x)}{-sin(x)} = \frac{\pi/2}{-\pi/2} = -1 [/tex]

    Edit: It is not pi/2 there it is supposed to be 1/-1
     
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