Two problems with limits.. absolute value

[lizzygarcia01]

I have two questions. They're not homework just looking at them and afraid they might be on my test tomorrow. I tried doing this one:

lim (2x+12)/(abs x+6) as x approaches -6. I tried doing it but I don't know if I can get rid of the absolute value. That's what I don't understand about this one.

The second is lim ((1/t) - (1/t^2+t)) as t approaches 0. I wasn't sure how to get rid of the t^2 +t. I'm scared this may be on the test tomorrow and need someone to explain it to me because I thought I understood everything until I seen his last examples.

 

[royhaas]

In the first case, evaluate the ratio for some numbers less than -6 and greater than -6. In the second case, simplify the expression first.

 

[galactus]

\(\displaystyle \frac{1}{t}-\frac{1}{t^{2}+t}=\frac{1}{t+1}\)

 

[lizzygarcia01]

Is this right then:
lim (2x+12)/(abs x+6) as x approaches -6
from the right:
2(x+6)/(abs x+6)=
2(x+6)/(x+6) = *Because the x will be positive since it's from the right*
2(x+6)/(x+6) =2

lim (2x+12)/(abs x+6) as x approaches -6
from the left:
2(x+6)/(abs x+6)=
2(x+6)/(-(x+ 6))=
*2(x+6)/(-1(x+6))=
2/(-1)=-2?

So the answer doesn't exist. Did I do that correctly? The part with the star is the one I think I might have done wrong.

((1/t) - (1/t^2+t)) as t approaches 0
((1/t)-(1/2(t(t+1))=
(((t+1/t(t+1))-(1/2(t(t+1))=
((t+1)-1)/(t(t+1))=
(t/t(t+1))=
1/(t+1)=
1/(0+1)=1

So the answer would be 1?

 

[galactus]

Yes and yes.