The function f(x)=(x-sinx)/x3 is continuous at x=0, then f(0) is
1The explanation to the above question has been given as :
lim(x→0)f(x)=f(0)
Applying LH rule (1-cosx)/2x2=sin2(x/2)/x2=1/4
However I solved it in the following way:
Applying LH rule 3 times= (1-cosx)/3x2= sinx/6x=cosx/6
hence lim(x→0)cosx/6=1/6
Could you please tell me what's wrong in this?
11This have to done by the following method
Use the expansion of Sin x =x-x3/3!..........
=+1/6
so ur answer is correct
1357you are absolutely right saket
there is a mistake in the solution
Applying LH rule (1-cosx)/2x2=sin2(x/2)/x2=1/4
instead of this it should have been
Applying LH rule (1-cosx)/3x2=2sin2(x/2)/3x2=1/6