Question #47a76

Mar 27, 2017

${\lim}_{x \to \frac{\pi}{2}} \frac{1 - {\sin}^{3} x}{{\cos}^{2} x} = \frac{3}{2}$

Explanation:

We have:

${\lim}_{x \to \frac{\pi}{2}} \frac{1 - {\sin}^{3} x}{{\cos}^{2} x}$

Now consider that in general:

$1 - {a}^{3} = \left(1 - a\right) \left({a}^{2} + a + 1\right)$

$1 - {a}^{2} = \left(1 - a\right) \left(1 + a\right)$

so that:

$\left(1 - {\sin}^{3} x\right) = \left(1 - \sin x\right) \left({\sin}^{2} x + \sin x + 1\right)$

${\cos}^{2} x = 1 - {\sin}^{2} x = \left(1 + \sin x\right) \left(1 - \sin x\right)$

So we can write the function as:

${\lim}_{x \to \frac{\pi}{2}} \frac{1 - {\sin}^{3} x}{{\cos}^{2} x} = {\lim}_{x \to \frac{\pi}{2}} \frac{\left(1 - \sin x\right) \left({\sin}^{2} x + \sin x + 1\right)}{\left(1 + \sin x\right) \left(1 - \sin x\right)}$

Simplifying the factor that is common to numerator and denominator:

${\lim}_{x \to \frac{\pi}{2}} \frac{1 - {\sin}^{3} x}{{\cos}^{2} x} = {\lim}_{x \to \frac{\pi}{2}} \frac{{\sin}^{2} x + \sin x + 1}{1 + \sin x} = \frac{3}{2}$

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