anonymous
  • anonymous
PLEASE I need help with two-sided limits
Mathematics
jamiebookeater
  • jamiebookeater
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.Sam.
  • .Sam.
What's the question?
.Sam.
  • .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
  • anonymous
hold on lemme write it out

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anonymous
  • anonymous
|dw:1329978579461:dw|
ggrree
  • 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
  • anonymous
Ugh my brain sucks right now good job ggree
anonymous
  • anonymous
if you plug in lim -3^(+) you will get the same answer which means the limit exists
anonymous
  • anonymous
thanks.....so what if there was an indeterminate like 40/0 or something, how will I go about solving that?
.Sam.
  • .Sam.
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anonymous
  • 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
  • anonymous
remember you are not using 0 but something infinitly close to zero
anonymous
  • anonymous
thus 4 or any real number would go into it infinity number of times
anonymous
  • anonymous
does that answer your question?
anonymous
  • anonymous
if you had -1/0^(-) = + infinity make sure to note that because a -/- = +
anonymous
  • anonymous
yea it kinda does....thanks
anonymous
  • anonymous
are you still confused with something?
anonymous
  • 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
  • anonymous
can I give you another problem to solve?
anonymous
  • anonymous
sure
anonymous
  • anonymous
|dw:1329979876435:dw|
anonymous
  • anonymous
but does that explain it but yeah post the question quick need to actually do my own school work lol
anonymous
  • anonymous
lol ok
anonymous
  • anonymous
not to rush you
anonymous
  • anonymous
|dw:1329980059353:dw|
anonymous
  • anonymous
for these type of questions you multiply by the conjugate so multiply the top and bottom by ((x)^(1/2) + 2)
anonymous
  • anonymous
did that help?
anonymous
  • anonymous
yes it did...thanks
anonymous
  • anonymous
no problem you should be pretty set for limits now if you understand limits going to infinity
anonymous
  • anonymous
i understand limits going to infinity but i need a resource for some two-sided limit problems. it's kind of confusing/
anonymous
  • 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
  • anonymous
thus the limit does not exist
anonymous
  • 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
  • anonymous
as the graph looks like |dw:1329981639188:dw|
anonymous
  • 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
  • anonymous
oh it makes sense to me now
anonymous
  • anonymous
ok good I was getting tired of explaing it although I should have did it right the first time

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