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?