Finding the Real Answer for C in a Limit Question

[zaddyzad]

Homework Statement

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

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.

 

[SammyS]

Homework Statement

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

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.

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

If not, then rationalize the numerator.

a³ - b³ = (a - b)(a²+ab+b²)

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

 

[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) ?

 

[Ray Vickson]

Homework Statement

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

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.

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

 

[HallsofIvy]

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) ?

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

 

[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 $(\heartsuit)$