Delta-epsilon limit of a rational function with a jump discontinuity

[NickBE]

Hello, I am trying to prove the following...

lim (x+3) $$\left|x+5\right|$$$$/x+5$$
x$$\rightarrow-5$$

from the left, L=+2
from the right, L=-2

I used delta-epsilon on the right hand limit and got $$\delta$$ = $$\epsilon$$

However, I'm not sure how to proceed when I get to this step while trying to prove the left hand limit:

$$\left|x+1\right|$$< $$\epsilon$$ if 0<$$\left|x+5\right|$$<$$\delta$$

 

[Dick]

If you are approaching from the left then |x+5|/(x+5)=(-1). So your expression f(x) becomes (-1)*(x+3). The quantity you want to prove less than epsilon is |L-f(x)|. I make that to be |2-(-1)(x+3)|=|x+5|.

 

[NickBE]

I think I see where you're going, but I want to make sure I have the right idea with the delta-epsilon definition.

lim f(x)=L
x->a

$$\left|f(x)-L\right|$$ <$$\epsilon$$ if 0<$$\left|x-a\right|$$< $$\delta$$

Is this the same definition that is used to find the one sided limits? If so, how is it manipulated to arrive at the conclusion you provided?

What I was trying to do as I attempted to prove the limit from the right:

$$\left|f(x)+2\right|$$ < $$\epsilon$$ if 0<$$\left|x+5\right|$$ < $$\delta$$

Which gave me:

$$\left|x+5\right|$$ <$$\epsilon$$ if 0<$$\left|x+5\right|$$ < $$\delta$$

and I called it a day on that one. But when I got to the second one, I tried using the same definition to try to prove the limit. I have a feeling that in the same format, it will give me an incorrect answer...where I am lacking understanding?