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

We'll verify if we get an indetermination. We'll
substitute x by the value of accumulation point.

lim
(2x-sin2x)/x^3 = (2*0 - sin 0)/0^3 = 0/0

Since the
indetermination is "0/0", we'll apply L'Hospital rule:

lim
(2x-sin2x)/x^3 = lim (2x-sin2x)'/(x^3)'

 lim
(2x-sin2x)'/(x^3)' = lim (2 - 2cos 2x)/3x^2

We'll
substitute x by the value of accumulation point.

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

Since the indetermination
is "0/0", we'll apply L'Hospital rule:

lim (2 - 2cos
2x)/3x^2 = lim (2 - 2cos 2x)'/(3x^2)'

 lim (2 - 2cos
2x)'/(3x^2)' = lim - (-4sin 2x)/(6x)

We'll substitute x by
the value of accumulation point.

lim (4sin 2x)/(6x) =
0/0

Since the indetermination is "0/0", we'll apply
again L'Hospital rule:

lim (4sin 2x)/(6x) = lim (4sin
2x)'/(6x)'

lim (4sin 2x)'/(6x)' = lim 8 cos
2x/6

We'll substitute x by the value of accumulation
point.

lim 8 cos 2x/6 = 8*cos 0/6 = 8*1/6 = 8/6 =
4/3

The value of the given limit, for
x->0, is: lim  (2x-sin2x)/x^3 = 4/3.