IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

nunknown91

see
on applying 'L' hospital
_{[x ]}

_{[ 0]} (sec² x - cosx)/(12sin² x . cosx)

= _{[x ]}

_{[0 ]} (1-cos³ x)/(3sin² 2x . cosx)

=_{[x ]}

_{[0 ]} (1-cos³ x)/(3sin2x.sin2x)

Also we have a formula that

for the following set

{ x, sinx, sin^-1 x, tanx, tan^-1 x, sinhx, tanhx, sinh^-1 x, tanh^-1 x }

following formula holds good

_{[x ]}

_{[0 ]} [1 - cos^p (ax)]/f(mx) . g(nx) = (p . a² )/2mn

Apply it and u will get it as 1/8

rate me if it was useful

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