What is the limit of function y=(1-cos^3 x)/(x^2) if x approaches to 0 ?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to find the value of lim x-->0 [(1- (cos x)^3)/(x^2)]

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

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

Again substitution of x = 0 yields 0/0, so repeat the above steps again

=> lim x-->0[(3*(cos x)^3-6*(cos x)*(sin x)^2)/2]

substitute x = 0

we get 3/2

The value of lim x-->0 [(1- (cos x)^3)/(x^2)] = 3/2

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We recognize to numerator a difference of 2 cubes and we'll apply the identity:

a^3 - b^3  =(a-b)(a^2 + ab + b^2)

1 - (cos x)^3 = (1 - cos x)[1 + cos x + (cos x)^2]

We'll re-write th limit:

lim [1 - (cos x)^3]/x^2 = lim (1 - cos x)[1 + cos x + (cos x)^2]/x^2

lim [1 - (cos x)^3]/x^2 = lim [(1 - cos x)/x^2]*lim[1 + cos x + (cos x)^2]

We'll evaluate the limit of each factor:

We'll substitute the numerator by the half angle formula:

1 - cos x = 2 [sin(x/2)]^2

Lim (1-cosx)/x^2 = Lim {2 [sin(x/2)]^2}/x^2

Lim {2 [sin(x/2)]^2}/x^2 = Lim  [sin(x/2)]^2/(x^2/2)

Lim [sin(x/2)]^2/(x^2/2) = lim[sin(x/2)/(x/2)]*lim[sin(x/2)/2*(x/2)]

Since the elementary limit is:

lim [sin f(x)]/f(x) = 1, if f(x) -> 0, we'll get:

lim (1-cosx)/x^2 = 1/2 (1)

We'll calculate the limit of the next factor:

lim[1 + cos x + (cos x)^2] = [1 + cos 0 + (cos 0)^2] = (1 + 1 + 1)=3 (2)

We'll multiply (1) by (2) and we'll get the limit of the function:

lim [1 - (cos x)^3]/x^2 = 3/2

The limit of the given function, if x approaches to 0, is: lim [1 - (cos x)^3]/x^2 = 3/2.

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