# Thread: Limits of absolute value function

1. ## Limits of absolute value function

Hello, I need some help with this

I need to find the limit of lim x->-2 2-|x| / 2+x

I need to evaluate the left and right handed limit. For the righthanded limit I plug in

lim -2+ 2-(x)/2+x,
now my problem is is to get something that can cancel out, so that I got the limit.

The lefthanded limit I did as following:
lim -2- 2-(-x)/2+x = lim -2- 2+x/2+x = 1

Thanks for any hint.

2. Originally Posted by DBA
Hello, I need some help with this

I need to find the limit of lim x->-2 2-|x| / 2+x

I need to evaluate the left and right handed limit. For the righthanded limit I plug in

lim -2+ 2-(x)/2+x,
now my problem is is to get something that can cancel out, so that I got the limit.

The lefthanded limit I did as following:
lim -2- 2-(-x)/2+x = lim -2- 2+x/2+x = 1

Thanks for any hint.
You don't need to calculate two sides of limit. Since when x->-2, you can assume that x<0.
so $\lim_{x\rightarrow -2}\frac{2-|x|}{2+x}=\lim_{x\rightarrow -2}\frac{2+x}{2+x}=1$

3. Sorry, but I do not understand that. How can I asume that x<0 in this case?
Can you try to explain this in a different way?
Thanks!

4. Originally Posted by DBA
Sorry, but I do not understand that. How can I asume that x<0 in this case?
Can you try to explain this in a different way?
Thanks!
Because the limit is only related to the values near -2, that is to say the value between $[-2+\delta,+\infty],[-\infty,-2-\delta],\forall\delta>0$is totally meaningless. So you can just consider the case when $x<0$.
Or you can refer the definition of the limit:
$\forall\epsilon>0,\exists\delta>0,\forall |x-(-2)|<\delta,|f(x)-limit|<\epsilon$. In this way, you may understand the "limit"better.
Since if $\delta_0$satisfy the definition for some $\epsilon>0$, then any $0<\delta<\delta_0$also satisfy the definition. So the limit is only related to the values near -2.

5. Thanks, I guess I was thinking in the wrong way. It took me a while to understand, but I guess what I did wrong was indeed not to see that in both cases x is smaller than zero and so I get 2+x in bith numerators.

I am still not sure if I understood the definition of the limit, because I do not know all these characters but I understand the problem to get an answer.