- #1

NickBE

- 2

- 0

lim (x+3) [tex]\left|x+5\right|[/tex][tex]/x+5[/tex]

x[tex]\rightarrow-5[/tex]

from the left, L=+2

from the right, L=-2

I used delta-epsilon on the right hand limit and got [tex]\delta[/tex] = [tex]\epsilon[/tex]

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

[tex]\left|x+1\right|[/tex]< [tex]\epsilon[/tex] if 0<[tex]\left|x+5\right|[/tex]<[tex]\delta[/tex]