Solve this limits question for me please.

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    I cant do it. Tell me the method, a link to a helpful website shall be useful.

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    (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
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    (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?
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    (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.
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    (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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    (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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