simple calculus // limit exist at x=-2?

Question

First graph has two arrows.
A line with an open circle at
(-2,2) going left forever horizontal.
another line with close circle at (-2,-2) going right horizontal forever.

second graph is just like the one above except that both start with open circles.

Third graph is strange because it does not have a circle but from that perticular point two arrows come out of it, with out circle or point. At (-2,2) one arrow goes right horizontal forever and the other one from again (-2,2) goes down at an angle hitting the x axis at (-4,0). It kinda looks like
3:40 on a clock. with (-2,2) in the middle.

Fourth graph has same starting point. there are two arrows from (-2,2).One is going down at an angle hitting the x axis at (-3,0). The other arrow is kinda strange because it starts from (-2,2) but then bends to the right alittle at (-1.5,2) then goes up till it hits the y axis at (0,6).

The question that i have is does the limit exist at x= -2

really need your help

Answer 1

A limit exists only of the limit approaching from the right equals the limit approaching from the left.

If the funtion is disjointed, no limit exists at that point.

so...

1. The limit does not exist at x=-2, since the function is disjointed at that point.
2. Same answer as 1.
3. Here the function is continuous so the limit at x=-2 exists.
4. Same answer as 3.

Answer 2

For graph1 the limit at x=-2 is -2

For graph 2 the limit does no exist at x = -2

For graphs 3 and 4 the limit at x= -2 is 2