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Ask iit jee aieee pet cbse icse state board experts Discussion Response Post to: Lt (x-->0) [1/Sin x][1/Sinx]-[1/Sinhx][1/Sinhx]=?
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iitkgp_bipin (6498)

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Olaaa!! Perrrfect answer. 1106  bad job dude!! I dont approve of this answer! 1  [1592 rates]

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i am assuming sinhx is hyperbolic sine of x.


 


sinhx = (ex - e-x) / 2


 


using series of ex and e-x and neglecting higher terms since x tends to 0.


 


(sinhx)2 = {(1 + x + x2/2) - (1 - x + x2/2)}2 / 4 = x2


 


so limit becomes


 



 


arrange it and use the series of sinx, you'll get 1/3.


 


 edited : Thanks allamraju, I did a slight calculation mistake. The answer is 2/3.


 


 


Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur

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