# [Math] Finding a limit with a cube

calculus

I am reviewing for a test and I can not figure this out.

$$\lim\limits_{h\to 0}\frac {(h-1)^3 + 1}{h}$$

I tried to multiply by the conjugate and that game me nothing sensible.

For $h\ne 0$: $${(h-1)^3+1\over h} ={(h^3-3h^2+3h-1)+1\over h }={h^2-3h+3 }.$$