Question

How do you evaluate `(sin^3x+cos^3x )/( cosx + sinx)` as x approaches `3pi/4`?

Answer 1

Use `a^3+b^3 = (a+b)(a^2-ab+b^2)` to simplify the ratio.

Explanation:

`lim_(xrarr3pi"/"4)(sin^3x+cos^3x)/(sinx+cosx) = lim_(xrarr3pi"/"4)((sinx+cosx)(sin^2x-sinxcosx+cos^2x))/(sinx+cosx)`

`= lim_(xrarr3pi"/"4)(sin^2x-sinxcosx+cos^2x)`

`= 1/2+1/2+1/2=3/2`