IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

uday_zingtudor

Got it buddy!!!

Now this is in indeterminate form.
Using l'hospital's rule,

in the numerator : 1/sqrt(1-x²) - 1

in the denominator: 3x²cosx - x³sinx

Again this is in indeterminate form and using l'hospitals rule again

in the numerator: [-1/2(1-x²)^3/2](-2x) = x/(1-x³)^3/2

in the denominator: 6xcosx - 3xsin²x - x²cosx - 3xsin²x
= 6xcosx - 6xsin²x - x²cosx

Now u can see, u can cancel 'x' in both numerator and denominator!!!

Substituting the limit, we get 1/6

~Cheerio!!!

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