Question #47a76

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Answer 1 · 2017-03-27T13:26:11

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

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]}