# Evaluate the limit of the function (2x-sin2x)/x^3; x-->0.

We have to determine lim x-->0 [(2x - sin 2x)/x^3]

If we substitute x = 0, we get the indeterminate form 0/0, so we use the l'Hopital's Rule and substitute the numerator and denominator with their derivatives

=> lim x-->0 [(2 - 2*cos 2x)/3*x^2]

substituting x = 0, we again get the form 0/0. Using the derivatives

=> lim x-->0 [(4*sin 2x)/6x]

substituting x = 0, we get 0/0 again, so we replace with the derivatives again

=> lim x-->0 [ 8*cos 2x/ 6]

substituting x = 0

=> 8/6

=> 4/3

The required value of lim x-->0 [(2x - sin 2x)/x^3] = 4/3

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