# Math Help - Help Solving the Limit of this equation (third root)

1. ## Help Solving the Limit of this equation (third root)

Find the solution for: [(1+cx)^1/3 - 1]/x as x approaches 0.

Hello, I've been trying to solve this problem for a little over a month without much progress..

Looking at it, the logical answer seems to be to get the (1+cx) out of it's third root form to cancel the constants and remove x, but I can't seem to do that w/o adding other similar terms. I've tried adding elements to the top to get a factorial out but that hasn't worked, I've tried multiplying by the positive counterpart ((1+cx)^1/3 + 1) to up the polynomial without luck, and lately I've been playing with natural logs to bring the 1/3 down; all to no avail.

I know the limit exists for sure, any help would be appreciated.

Thanks,
Rhek

2. ## Re: Help Solving the Limit of this equation (third root)

Have you tried L'hospital's rule?

L'Hôpital's rule - Wikipedia, the free encyclopedia

3. ## Re: Help Solving the Limit of this equation (third root)

I've never like using "L'Hopital" if a limit could be done without Calculus.

Recall that $x^3- y^3= (x- y)(x^2+ xy+ y^2)$

Taking y= 1,
$x^3- 1= (x- 1)(x^2+ x+ 1)$

Replacing x by $a^{1/3}$, $x^3= a$ and $x^2= a^{2/3}$ and $a- 1= (a^{1/3}- 1)(a^{2/3}+ a^{1/3}+ 1)$

That is, $((1+cx)^{1/3}- 1)/x$ can be written as
$(1+cx- 1)/[x((1+cx)^{2/3}+ (1+ cx)^{1/3}+ 1)]= cx/[x((1+cx)^{2/3}+ (1+ cx)^{1/3}+ 1)]$ $= c/((1+cx)^{2/3}+ (1+ cx)^{1/3}+ 1)$

And, the limit of that, as x goes to 0, is easy.

4. ## Re: Help Solving the Limit of this equation (third root)

Thanks for both the responses guys. I should have stipulated that the problem is meant to be done w/o the use of derivatives, but now I/ve got both solutions. Thanks so much for the help, been scratching my head over that one for a long time.

-Rhek