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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.

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    Use identity $(a-b)^3=a^3-3a^2b+3ab^2-b^3$2012-02-06

2 Answers 2

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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 }. $

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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 $3$.

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    @Michael I've also seen many calculus problems giving limits that can be manipulated to this $f$orm, expecting the student to recognize and apply the de$f$inition. It depends on what point in the class this is being asked.2012-02-06