# Limit question, evalute for C

1. Feb 20, 2013

### zaddyzad

1. The problem statement, all variables and given/known data

lim $\frac{\sqrt[3]{1+cx}-1}{x}$
x$\rightarrow$0

3. The attempt at a solution

I put 0 in for x, which lead to 0/0 therefore I know I know I need to modify to get a real answer. I have no idea where to start.

2. Feb 20, 2013

### SammyS

Staff Emeritus
Are you allowed to use L'Hôpital's rule ?

If not, then rationalize the numerator.

a3 - b3 = (a - b)(a2+ab+b2)

So that $\displaystyle \ \ (\sqrt[3]{s}-\sqrt[3]{t})((\sqrt[3]{s})^2+\sqrt[3]{s}\sqrt[3]{t}+(\sqrt[3]{t})^2)=s-t$

3. Feb 20, 2013

### zaddyzad

So it's like multiply by a conjugate for square root rationals, but for a cube root??

Equivalent to x^2-y^2=(x-y)(x+y) ???

4. Feb 20, 2013

### Ray Vickson

Do you know about Taylor series? If so, just take the first few terms of the expansion of $(1+cx)^{1/3}$ about $x=0$.

5. Feb 20, 2013

### HallsofIvy

Staff Emeritus
Yes, but x^3- y^3= (x- y)(x^2+ xy+ y^2).

6. Feb 20, 2013

### Mathitalian

or you can use the limit:

$\displaystyle\lim_{\begin{matrix}f(x)\to 0\\\mbox{when }x\to x_0\end{matrix}}\frac{(1+f(x))^\alpha-1}{f(x)}= \alpha\qquad (\heartsuit)$

You have just to multiply and divide by $c\ne 0$ and use $(\heartsuit)$

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