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.
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.
Try expanding it out.
For $h\ne 0$: $ {(h-1)^3+1\over h} ={(h^3-3h^2+3h-1)+1\over h }={h^2-3h+3 }. $
You may also want to note that this is equivalent to the derivative of $x^3$ evaluated at $-1$, using the definition
f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}
with $f(x)=x^3$.
So since f'(-1)=3(-1)^2=3, the limit is also $3$.