Find the limit of the function algebraically.

-6
-3
Does not exist
0
Im kind of confused on how to do this exactly. Could someone maybe walk me through the steps?

Question

Find the limit of the function algebraically. 


 	
	-6
	-3
	Does not exist
	0
Im kind of confused on how to do this exactly. Could someone maybe walk me through the steps?

anonymous · Answer

Oh hold on you cant see the equation

anonymous · Answer

lim of x approaching -3, (x^2-9)/(x^3+3)

anonymous · Answer

where you see an x, put a 3
that is all

anonymous · Answer

negative 3 or just 3

anonymous · Answer

if you get $\frac{0}{0}$ then you have more work to do 
oh right, -3 what you said

anonymous · Answer

but in this case you get  0 over something that is not zero, answer is just zero

anonymous · Answer

So i did that and got -18/-24

anonymous · Answer

you dont get 0 on top because its -3^2-9 which would be -18

anonymous · Answer

oh no

anonymous · Answer

$(-3)^2=9 $ not $-3^2=-9$

anonymous · Answer

OHH my gosh. I keep making such dumb mistakes. duhhh. thank you!!

anonymous · Answer

lol yw

anonymous · Answer

Im having trouble with limits. Would you like to help me?

anonymous · Answer

https://partners-bluemouse.brainhoney.com/Resource/25952867,444,0,0,1A,2,0,0/Assets/46194_50f975f1/0906_g11_q3.gif heres the equation

anonymous · Answer

here are the possible answers 
The limit does not exist.
	-1
	-4
	7

anonymous · Answer

I graphed it but how do i know when the limit exists.? Do they need to have the same point when they approach -4?

anonymous · Answer

limit from the left is not equal to the limit from the right, so it does not exist

anonymous · Answer

yeah right, they need to be teh same

anonymous · Answer

Okay, thank you tons!!

anonymous · Answer

yw

anonymous · Answer

tons