- anonymous

PLEASE I need help with two-sided limits

- jamiebookeater

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- .Sam.

What's the question?

- .Sam.

If you want to know what's two-sided limits is,
A two sided limit is just the regular limit you see denoted by the lim as x approaches some value of a function. It means that if you approached that value from both sides of the graph you would arrive at the same place.
A one sided limit means if you approached the graph from one particular side (from the left or from the right) you would get different values. If the limit from the left does not equal the limit from the right side of the graph, the overall limit does not exist.
For example, a function such as y = x, for x > 0 cannot have a negative x-value. So the limit of this function from the left would not exist because the graph doesn't exist there, but the limit from the right would be 0.

- anonymous

hold on lemme write it out

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## More answers

- anonymous

|dw:1329978579461:dw|

- ggrree

factor the top and the bottom to get \[x(x+3) \over (x+3)(x-3)\]
cancelling gives you:
\[x \over x-3\]
now you can simply plug in -3.

- anonymous

Ugh my brain sucks right now good job ggree

- anonymous

if you plug in lim -3^(+) you will get the same answer which means the limit exists

- anonymous

thanks.....so what if there was an indeterminate like 40/0 or something, how will I go about solving that?

- .Sam.

##### 1 Attachment

- anonymous

4/0^(+)= + infinity
4/0^(-) = - infinity
If you get 0/0 you did something wrong unless the prof teaches you L'hospitols rule

- anonymous

remember you are not using 0 but something infinitly close to zero

- anonymous

thus 4 or any real number would go into it infinity number of times

- anonymous

does that answer your question?

- anonymous

if you had -1/0^(-) = + infinity
make sure to note that because a -/- = +

- anonymous

yea it kinda does....thanks

- anonymous

are you still confused with something?

- anonymous

if you take the limit from the 0+ of 1/x you get +infinity
and if you take the limit from 0- you get -infinity
I think I made a mistake with the graph, but yeah the out put as you approach from the right is always increasing
and the output from the left is always decreasing

- anonymous

can I give you another problem to solve?

- anonymous

sure

- anonymous

|dw:1329979876435:dw|

- anonymous

but does that explain it but yeah post the question quick need to actually do my own school work lol

- anonymous

lol ok

- anonymous

not to rush you

- anonymous

|dw:1329980059353:dw|

- anonymous

for these type of questions you multiply by the conjugate so multiply the top and bottom by
((x)^(1/2) + 2)

- anonymous

did that help?

- anonymous

yes it did...thanks

- anonymous

no problem you should be pretty set for limits now if you understand limits going to infinity

- anonymous

i understand limits going to infinity but i need a resource for some two-sided limit problems. it's kind of confusing/

- anonymous

did my 1/x example not help?
1/0- = something like 1/-0.0000001
where as
1/0+ = something like 1/+0.000001
Only these numbers are infinitely small so we end up with
lim 1/x = infinity
x-> 0+
lim 1/x = -infinity
x -> 0-
Look at the graph of 1/x
|dw:1329981425836:dw|
Notice how the output is forever decreasing from the left as you approach zero
whereas the output is forever increasing from the right as you approach zero

- anonymous

thus the limit does not exist

- anonymous

if you took the limit of -1/x
you would get the opposite if you took limits at both side
so
0+ = -infinity
0- = +infinity

- anonymous

as the graph looks like
|dw:1329981639188:dw|

- anonymous

if you took the limit of 0+ and 0- of x
you would end up with
0 and 0 which means that the limit exists

- anonymous

oh it makes sense to me now

- anonymous

ok good I was getting tired of explaing it although I should have did it right the first time

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