The Student Room Group

Solve this limits question for me please.

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

Reply 1
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 00 \frac{0}{0} or \frac{\infty}{\infty} form and use the L'Hospital rule.

1, limx1+lnxcosπx2limx1+sinπx2lim_{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 limx1+sinπx2lim_{x\rightarrow 1^+} sin \frac {\pi x}{2} will canceled out
Reply 2
Original post by ztibor
With identities arrange the limit to 00 \frac{0}{0} or \frac{\infty}{\infty} form and use the L'Hospital rule.

1, limx1+lnxcosπx2limx1+sinπx2lim_{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 limx1+sinπx2lim_{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?
(edited 11 years ago)
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.
Reply 4
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?


(cosπx2)=sinπx2π2\displaystyle \left (cos \frac{\pi x}{2}\right )' =-sin\frac{\pi x}{2} \cdot \frac {\pi}{2}
Reply 5
Original post by ztibor


(cosπx2)=sinπx2π2\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...

Quick Reply

Latest