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

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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**

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.**