finding the limit where the equation contains a cube root

G

Guest

Guest
I'm having trouble again. I'm supposed to find the limit, but the cube root seems to be messing everything up.
The question:
lim(x->infinity) (x^(1/3) - (x/3))

The answer is supposed to be -infinity, but I've gotten 0, infinity, and at one time an undefined half way through.
I know I should use (a-b)(a^2 +ab+b^2)=a^3 -b^3 however I seem to be messing up somewhere along the way. If someone would direct me to some directions on how to set up the TeX stuff, I'd write it all out, but it would get very messy, very fast if I tried to without it. In fact, it's pretty ugly-looking on this sheet of paper beside me.
 
G

Guest

Guest
Never mind! I found it at the top of the page. Give me some time, but here's what I've typed out so far:
\(\displaystyle \L (a - b)(a^2 + ab + b^2) = a^3 - b^3\)
\(\displaystyle \L If a = \sqrt[3]{x} and b = x/3\)
\(\displaystyle \L then \frac{(\sqrt[3]{x})^3 - (x/3)^3}{(\sqrt[3]{x})^2 + \sqrt[3]{x}(x/3) + (x/3)^2}\)
 

galactus

Super Moderator
Staff member
Joined
Sep 28, 2005
Messages
7,216
Like this:

\(\displaystyle \L\\\lim_{x\to\infty}[x^{\frac{1}{3}}-\frac{x}{3}]\)

\(\displaystyle \L\\\lim_{x\to\infty}[3x^{\frac{1}{3}}-x]\)

Divide by \(\displaystyle x^{\frac{1}{3}}\)

\(\displaystyle \L\\\lim_{x\to\infty}[3-x^{\frac{2}{3}}]\)
 
G

Guest

Guest
Thank you! That's so simple and I probably would never have gotten there.
At least I learned some of the TeX stuff while I was at it.
 
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