Question

Question #47a76

Answer 1

`lim_(x->pi/2) (1-sin^3x)/(cos^2x) = 3/2`

Explanation:

We have:

`lim_(x->pi/2) (1-sin^3x)/(cos^2x)`

Now consider that in general:

`1-a^3 = (1-a)(a^2+a+1)`

`1-a^2 = (1-a)(1+a)`

so that:

`(1-sin^3x) = (1-sinx)(sin^2x+sinx+1)`

`cos^2x = 1-sin^2x = (1+sinx)(1-sinx)`

So we can write the function as:

`lim_(x->pi/2) (1-sin^3x)/(cos^2x) = lim_(x->pi/2) ((1-sinx)(sin^2x+sinx+1))/((1+sinx)(1-sinx))`

Simplifying the factor that is common to numerator and denominator:

`lim_(x->pi/2) (1-sin^3x)/(cos^2x) = lim_(x->pi/2) (sin^2x+sinx+1)/(1+sinx) =3/2`

graph{(1-(sinx)^3)/((cosx)^2) [-2.167, 2.833, -0.13, 2.37]}