Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

PLEASE I need help with two-sided limits

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

What's the question?
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.
hold on lemme write it out

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question