Lim,integral,continuous graph

[spezialize]

Hey guys. I am having an issue with some graph and telling if the statements are true/false.

that is a crappy drawing.. and here is what i am trying to find out. if you could kind of explain a little bit to so I can understand it somewhat better. Thats right I don't just want answers for hw, but want to understand why.

lim f(x) =1 as x approaches -2 from the left
lim f(x) =1 as x approaches -2 from the right
lim f(x) Doesnt exist as x approaches -1

integral { 1 to 3} f(x) dx = 1
integral {0 to 3} f(x) dx = 3/2
function is continuous at x=0
function is continuous at x=-1
function is continuous at x=2
function is continuous at x=-1

Thank you for trying to help me understand calculus through graphs[/tex]

 

[Gene]

There is a good bit of guessing but...
http://img226.imageshack.us/my.php?image=calc8xf.png

It seems to be intended to be made up of six straight lines.

lim f(x) =1 as x approaches -2 from the left
The line goes toward 0, not 1
lim f(x) =1 as x approaches -2 from the right
The line does go toward 1 but f(-2) = 2, not 1
lim f(x) Doesnt exist as x approaches -1
The limit exists 'cause it can get as close to 0 as you want but doesn't reach it.
integral { 1 to 3} f(x) dx = 1
That is a triangle with a base = 2 and height = 1
integral {0 to 3} f(x) dx = 3/2
The area from x = 0 to 1 is negative and equal to the area from x=1 to 2 so they cancel leaving the triangle from 2 to 3 as the integral.
function is continuous at x=0
It approaches -1 from both directions and is defined as -1 at x=0
function is continuous at x=-1
It is not defined at x = -1
function is continuous at x=2
same as x=0
function is continuous at x=-1
I haven't changed my mind

You might want to look at
http://mathworld.wolfram.com/ContinuousFunction.html

 

[spezialize]

thank you. also the link was helpful!