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.Sam. Group TitleBest ResponseYou've already chosen the best response.0
What's the question?
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.0
If you want to know what's twosided 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 xvalue. 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.
 2 years ago

aroberts47 Group TitleBest ResponseYou've already chosen the best response.0
hold on lemme write it out
 2 years ago

aroberts47 Group TitleBest ResponseYou've already chosen the best response.0
dw:1329978579461:dw
 2 years ago

ggrree Group TitleBest ResponseYou've already chosen the best response.2
factor the top and the bottom to get \[x(x+3) \over (x+3)(x3)\] cancelling gives you: \[x \over x3\] now you can simply plug in 3.
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
Ugh my brain sucks right now good job ggree
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
if you plug in lim 3^(+) you will get the same answer which means the limit exists
 2 years ago

aroberts47 Group TitleBest ResponseYou've already chosen the best response.0
thanks.....so what if there was an indeterminate like 40/0 or something, how will I go about solving that?
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
4/0^(+)= + infinity 4/0^() =  infinity If you get 0/0 you did something wrong unless the prof teaches you L'hospitols rule
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
remember you are not using 0 but something infinitly close to zero
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
thus 4 or any real number would go into it infinity number of times
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
does that answer your question?
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
if you had 1/0^() = + infinity make sure to note that because a / = +
 2 years ago

aroberts47 Group TitleBest ResponseYou've already chosen the best response.0
yea it kinda does....thanks
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
are you still confused with something?
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

aroberts47 Group TitleBest ResponseYou've already chosen the best response.0
can I give you another problem to solve?
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
dw:1329979876435:dw
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
but does that explain it but yeah post the question quick need to actually do my own school work lol
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
not to rush you
 2 years ago

aroberts47 Group TitleBest ResponseYou've already chosen the best response.0
dw:1329980059353:dw
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
for these type of questions you multiply by the conjugate so multiply the top and bottom by ((x)^(1/2) + 2)
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
did that help?
 2 years ago

aroberts47 Group TitleBest ResponseYou've already chosen the best response.0
yes it did...thanks
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
no problem you should be pretty set for limits now if you understand limits going to infinity
 2 years ago

aroberts47 Group TitleBest ResponseYou've already chosen the best response.0
i understand limits going to infinity but i need a resource for some twosided limit problems. it's kind of confusing/
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
thus the limit does not exist
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
as the graph looks like dw:1329981639188:dw
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

aroberts47 Group TitleBest ResponseYou've already chosen the best response.0
oh it makes sense to me now
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
ok good I was getting tired of explaing it although I should have did it right the first time
 2 years ago
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