# Thread: Another Limit problem :X

1. ## 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
Please give me a helping hand! Thank you!

2. 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
Please give me a helping hand! Thank you!

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.

3. 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?

4. Originally Posted by greenbee
Please tell me that it's 16?
No, it's not 16.

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

5. 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?

6. Originally Posted by greenbee
It doesn't go to 0
It goes to like -Infinity
So the answer is -Infinity?
Yes, $\lim_{x\to0^-}\frac{16}{x}=-\infty$.

7. Originally Posted by Chris L T521
Yes, $\lim_{x\to0^-}\frac{16}{x}=-\infty$.
YAY thank you so much!!! (Thanks thanks thanks button)!!!