Solve this limits question for me please.

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  1. zedeneye1's Avatar
    1 16-04-2012  07:13
    • Exalted and Worshipped Member
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    Solve this limits question for me please.
    I cant do it. Tell me the method, a link to a helpful website shall be useful.

  2. ztibor's Avatar
    2 16-04-2012  07:41
    • Peer Of The TSR Realm
    • Location: Hungary
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    Re: Solve this limits question for me please.
    (Original post by zedeneye1)
    I cant do it. Tell me the method, a link to a helpful website shall be useful.

    With identities arrange the limit to \frac{0}{0} or \frac{\infty}{\infty} form and use the L'Hospital rule.

    1, lim_{x\rightarrow 1^+} \frac {lnx}{cos \frac{\pi x}{2}} \cdot lim_{x\rightarrow 1^+} sin \frac {\pi x}{2}

    2. Use the L"Hospital to the first limit
    the lim_{x\rightarrow 1^+} sin \frac {\pi x}{2} will canceled out
  3. zedeneye1's Avatar
    3 16-04-2012  10:33
    • Exalted and Worshipped Member
    • Posts: 1,307
    Re: Solve this limits question for me please.
    (Original post by ztibor)
    With identities arrange the limit to \frac{0}{0} or \frac{\infty}{\infty} form and use the L'Hospital rule.

    1, lim_{x\rightarrow 1^+} \frac {lnx}{cos \frac{\pi x}{2}} \cdot lim_{x\rightarrow 1^+} sin \frac {\pi x}{2}

    2. Use the L"Hospital to the first limit
    the lim_{x\rightarrow 1^+} sin \frac {\pi x}{2} will canceled out
    oh yeah, forgot i cud use L'hospital rule.....thanks.

    but wait, doing it ur way wud give answer=1 which is not the answer...
    so wat now?
    Last edited by zedeneye1; 16-04-2012 at 11:34.
  4. ben-smith's Avatar
    4 16-04-2012  12:48
    • Overlord in Training
    • Location: Hilbert Space
    • Posts: 2,367
    Re: Solve this limits question for me please.
    (Original post by zedeneye1)
    oh yeah, forgot i cud use L'hospital rule.....thanks.

    but wait, doing it ur way wud give answer=1 which is not the answer...
    so wat now?
    you haven't used the chain rule correctly.
  5. ztibor's Avatar
    5 16-04-2012  14:17
    • Peer Of The TSR Realm
    • Location: Hungary
    • Posts: 1,537
    Re: Solve this limits question for me please.
    (Original post by zedeneye1)
    oh yeah, forgot i cud use L'hospital rule.....thanks.

    but wait, doing it ur way wud give answer=1 which is not the answer...
    so wat now?
    \displaystyle \left (cos \frac{\pi x}{2}\right )' =-sin\frac{\pi x}{2} \cdot \frac {\pi}{2}
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  6. zedeneye1's Avatar
    6 16-04-2012  18:48
    • Exalted and Worshipped Member
    • Posts: 1,307
    Re: Solve this limits question for me please.
    (Original post by ztibor)

    \displaystyle \left (cos \frac{\pi x}{2}\right )' =-sin\frac{\pi x}{2} \cdot \frac {\pi}{2}
    thanks i got it now...

    thank you very much, both of you guys...
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