## aroberts47 Group Title PLEASE I need help with two-sided limits 2 years ago 2 years ago

1. .Sam.

What's the question?

2. .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.

3. aroberts47

hold on lemme write it out

4. aroberts47

|dw:1329978579461:dw|

5. 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.

6. cuddlepony

Ugh my brain sucks right now good job ggree

7. cuddlepony

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

8. aroberts47

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

9. .Sam.

10. cuddlepony

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

11. cuddlepony

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

12. cuddlepony

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

13. cuddlepony

14. cuddlepony

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

15. aroberts47

yea it kinda does....thanks

16. cuddlepony

are you still confused with something?

17. cuddlepony

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

18. aroberts47

can I give you another problem to solve?

19. cuddlepony

sure

20. cuddlepony

|dw:1329979876435:dw|

21. cuddlepony

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

22. aroberts47

lol ok

23. cuddlepony

not to rush you

24. aroberts47

|dw:1329980059353:dw|

25. cuddlepony

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

26. cuddlepony

did that help?

27. aroberts47

yes it did...thanks

28. cuddlepony

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

29. aroberts47

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

30. cuddlepony

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

31. cuddlepony

thus the limit does not exist

32. cuddlepony

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

33. cuddlepony

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

34. cuddlepony

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

35. aroberts47

oh it makes sense to me now

36. cuddlepony

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