LCD: Limits Continuity and Differentiability

if function is continuous at zero, so it has to be equal to its limit!

so we need to find

limx->0 f(x)

now clearly at zero it is 0/0 form

we apply LH rule..

limx->0 f(x) = (3cos3x+2Acos2x+bcosx)/5x⁴

This is defined. So numerator has to be zero! othewise it will not have limit

3cos0+2Acos0+bcos0=0 thus, 3+2A+B=0

again, we differentiate the above to find the lim (Apply LH rule)

limx->0 f(x) = (-9sin3x-4Asin2x-bsinx)/20x³

This is of the form 0/0.. so we get nothing much here.. we will have to differentiate again! (I mean apply LH Rule)

limx->0 f(x)= (-27cos3x-8Acos2x-Bcosx)/60x²

at x=0 , this = (-27-8A-B) = 0

thus.. we have 2 equation. solve them simultaneously ...
3+2A+B=0 and -27-8A-B = 0
u will get the answer :)