How do you find the Limit of #(sin^3 x) / (sin x - tan x)# as x approaches 0?

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Answer 1 · 2016-12-23T09:45:29

#lim_(x->0)frac (sin^3x) (sinx-tanx)=-2#

Start from:

#f(x) = frac (sin^3x) (sinx-tanx)#

Divide numerator and denominator by #sinx#:

#f(x) = frac (sin^2x) (1-1/cosx)=frac (sin^2xcosx) (cosx-1)#

Now divide by #x^2# above and below the line:

#f(x) = -cosx frac ((sinx/x)^2) ((1-cosx)/x^2)#

We know that:

#lim_(x->0) sinx/x =1#

#lim_(x->0) (1-cosx)/(x^2) =1/2#

so:

#lim_(x->0)frac (sin^3x) (sinx-tanx)=lim_(x->0)-cosx frac ((sinx/x)^2) ((1-cosx)/x^2)= (-1)* 1^2/(1/2)=-2#