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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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    True, but the main reason for studying this kind of limit is in order to _prove_ things like $(d/dx) x^3=3x^2$, and then you can't assume that's _already_ known.2012-02-06
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    @Michael I've also seen many calculus problems giving limits that can be manipulated to this form, expecting the student to recognize and apply the definition. It depends on what point in the class this is being asked.2012-02-06