# Limit of Absolute Value of X

1. Sep 21, 2008

### ne12o

Here, it says that for the limit f(x) = |x| / x,

|x| = { x, x > 0
-x, x < 0 }

What I don't undestand is why is |x| = -x for values under zero? Isn't the absolute value for negative values just x and not -x?

thanks.

EDIT: I don't want to start a new thread, but I got stuck on this next question :(

Lim x approaching -1 of
3(1-x^2) / x^3 + 1

I tried multiply the equation by x^3 - 1 / x^3 - 1
and I ended up with 0 / -2.

Thanks

Last edited: Sep 21, 2008
2. Sep 21, 2008

### morphism

Example: |-7| = -(-7) = 7.

Do you see why we need the minus sign?

3. Sep 21, 2008

### ne12o

Wow!!1
Thanks a lot! I can't believe I cannot see such a simple logic...

4. Sep 21, 2008

### haushofer

Ofcourse, what you could do is, by knowing in forhand the limit, to use the delta-epsilon method. But that would become quite messy, I think.

Your strategy of multiplying the limit with 1 with apropriate numerator and denominator is clever, and is often useful when you deal with square roots.

Here however, you see that the denominator and numerator have the same limit: 0. That's an indication to use L'hopital. As far as I can see your limit is equal to

$$lim_{x \rightarrow -1} \frac{3-3x^{2}}{1+x^{3}} = lim_{x \rightarrow -1}\frac{-6x}{3x^{2}} = lim_{x \rightarrow -1} \frac{-2}{x} = 2$$

5. Sep 21, 2008

### ne12o

Thanks for the help! I was stuck on this equation for a while.
It's been about a year since I've even touched Calculus... I'm extremely rusty on the basics, and now I have to relearn everything!