Question

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Answer 1

start with
\[\frac{-2(x+h)^3+2x^3}{h}\] and then expand the cube

Answer 2

you get
\[\frac{-2(x^2+3x^2h+3xh^2+h^3)+2x^3}{h}\]

Answer 3

distribute the -2 to get
\[\frac{-2x^3-6x^2h-6xh^2-2h^3+2x^3}{h}\]

Answer 4

\[-2x^3+2x^3=0\] so your fraction is
\[\frac{-6x^2h-6xh^2-2h^3}{h}=\frac{h(-6x^2-6xh-2h^2)}{h}\]
\[=-6x^2-6xh-2h^2\] and take the limit as h goes to zero, which just means replacing h by zero and you get
\[-6x^2\]

Answer 5

Thank you so much!