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[Post New]8 Apr 2008 22:37:58 IST
    Subject: Re:solve this limit
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ramkumar_november (1270)

Blazing goIITian

Olaaa!! Perrrfect answer. 230  [290 rates]
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l   =  \lim_{x\to 0} \bigg(\frac{sinx}{x}\bigg)^{\frac{sinx}{x-sinx}}
 
l  =  \lim_{x\to 0} \bigg(1+\frac{sinx-x}{x}\bigg)^{\frac{sinx}{x-sinx}}
 
l  = \lim_{x\to 0} \bigg(1+\frac{sinx-x}{x}\bigg)^{\frac{x}{sinx-x}.\frac{-sinx}{x}}
 
l  =  e^{-1}
 
method already given by sandeepramesh.....
i just latexified it......
 
hope you are clear.........
 this reply: 5 points  (with 1 Olaaa!! Perrrfect answer.   in 1 votes )   [?]
 
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