what is the answer to the limit problem : lim as x goes to zero ; 4 (sinx cosx- sin x)/ x^2?

Question

please help!

Answer 1

shasjing is correct but perhaps a little short on the explanation:

4(sinx*cosx-cosx)=4*sinx*(cosx-1) and the Taylor series are sin(x)=x+ other stuff while cos(x)=1-x^2/2+other stuff so 4*sinx*(cosx-1) is about 4*x*-x^2/2 which is -2x^3. then -2x^3/x^2 = -2x -> 0 as x -> 0.

If you don't use Taylor's series you are going to have to do L'Hopital rule,,,,,,,,,,

Answer 2

The answer is zero because sinx cosx - sin x has higher order of zero than x^2.

Answer 3

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Answer 4

Thus 2(sin2x –2sinx) / x^2 ={l’hopital} = (2cos2x –2cosx) / x = {l’hopital} = -4sin2x +2sinx =0;