# Another Limit problem :X

• September 20th 2009, 08:11 PM
greenbee
Another Limit problem :X
Do x is NONE? Since 8/0 = undefined if I plug 0 into x? Agree? Just want to make sure. I am not good with math (Surprised)
(Happy) Please give me a helping hand! Thank you!

http://i38.tinypic.com/a09hkk.jpg
• September 20th 2009, 08:17 PM
Chris L T521
Quote:

Originally Posted by greenbee
Do x is NONE? Since 8/0 = undefined if I plug 0 into x? Agree? Just want to make sure. I am not good with math (Surprised)
(Happy) Please give me a helping hand! Thank you!

http://i38.tinypic.com/a09hkk.jpg

Recall that $\left|x\right|=\left\{\begin{array}{rl}x & x\geq 0\\ {\color{red}-x} & {\color{red}x<0}\end{array}\right.$

So in your case,

$\lim_{x\to0^-}\left(\frac{8}{x}-\frac{8}{\left|x\right|}\right)=\lim_{x\to0^-}\left(\frac{8}{x}+\frac{8}{x}\right)=\lim_{x\to 0^-}\frac{16}{x}$

I'm sure you can take it from here.
• September 20th 2009, 08:32 PM
greenbee
Quote:

Originally Posted by Chris L T521
Recall that $\left|x\right|=\left\{\begin{array}{rl}x & x\geq 0\\ {\color{red}-x} & {\color{red}x<0}\end{array}\right.$

So in your case,

$\lim_{x\to0^-}\left(\frac{8}{x}-\frac{8}{\left|x\right|}\right)=\lim_{x\to0^-}\left(\frac{8}{x}+\frac{8}{x}\right)=\lim_{x\to 0^-}\frac{16}{x}$

I'm sure you can take it from here.

Please tell me that it's 16? (Thinking)
• September 20th 2009, 08:42 PM
Prove It
Quote:

Originally Posted by greenbee
Please tell me that it's 16? (Thinking)

No, it's not 16.

Try graphing $f(x) = \frac{16}{x}$ and see what happens when $x \to 0$ from the left...
• September 20th 2009, 08:49 PM
greenbee
Quote:

Originally Posted by Prove It
No, it's not 16.

Try graphing $f(x) = \frac{16}{x}$ and see what happens when $x \to 0$ from the left...

It doesn't go to 0
It goes to like -Infinity
So the answer is -Infinity? (Thinking)
• September 20th 2009, 08:50 PM
Chris L T521
Quote:

Originally Posted by greenbee
It doesn't go to 0
It goes to like -Infinity
So the answer is -Infinity? (Thinking)

Yes, $\lim_{x\to0^-}\frac{16}{x}=-\infty$.
• September 20th 2009, 08:52 PM
greenbee
Quote:

Originally Posted by Chris L T521
Yes, $\lim_{x\to0^-}\frac{16}{x}=-\infty$.

YAY thank you so much!!! (Happy) (Thanks thanks thanks button)!!!