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aroberts47

PLEASE I need help with two-sided limits

  • 2 years ago
  • 2 years ago

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  1. .Sam.
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    What's the question?

    • 2 years ago
  2. .Sam.
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    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.

    • 2 years ago
  3. aroberts47
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    hold on lemme write it out

    • 2 years ago
  4. aroberts47
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    |dw:1329978579461:dw|

    • 2 years ago
  5. ggrree
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    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.

    • 2 years ago
  6. cuddlepony
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    Ugh my brain sucks right now good job ggree

    • 2 years ago
  7. cuddlepony
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    if you plug in lim -3^(+) you will get the same answer which means the limit exists

    • 2 years ago
  8. aroberts47
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    thanks.....so what if there was an indeterminate like 40/0 or something, how will I go about solving that?

    • 2 years ago
  9. .Sam.
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    • 2 years ago
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  10. cuddlepony
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    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
  11. cuddlepony
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    remember you are not using 0 but something infinitly close to zero

    • 2 years ago
  12. cuddlepony
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    thus 4 or any real number would go into it infinity number of times

    • 2 years ago
  13. cuddlepony
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    does that answer your question?

    • 2 years ago
  14. cuddlepony
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    if you had -1/0^(-) = + infinity make sure to note that because a -/- = +

    • 2 years ago
  15. aroberts47
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    yea it kinda does....thanks

    • 2 years ago
  16. cuddlepony
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    are you still confused with something?

    • 2 years ago
  17. cuddlepony
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    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
  18. aroberts47
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    can I give you another problem to solve?

    • 2 years ago
  19. cuddlepony
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    sure

    • 2 years ago
  20. cuddlepony
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    |dw:1329979876435:dw|

    • 2 years ago
  21. cuddlepony
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    but does that explain it but yeah post the question quick need to actually do my own school work lol

    • 2 years ago
  22. aroberts47
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    lol ok

    • 2 years ago
  23. cuddlepony
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    not to rush you

    • 2 years ago
  24. aroberts47
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    |dw:1329980059353:dw|

    • 2 years ago
  25. cuddlepony
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    for these type of questions you multiply by the conjugate so multiply the top and bottom by ((x)^(1/2) + 2)

    • 2 years ago
  26. cuddlepony
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    did that help?

    • 2 years ago
  27. aroberts47
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    yes it did...thanks

    • 2 years ago
  28. cuddlepony
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    no problem you should be pretty set for limits now if you understand limits going to infinity

    • 2 years ago
  29. aroberts47
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    i understand limits going to infinity but i need a resource for some two-sided limit problems. it's kind of confusing/

    • 2 years ago
  30. cuddlepony
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    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
  31. cuddlepony
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    thus the limit does not exist

    • 2 years ago
  32. cuddlepony
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    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
  33. cuddlepony
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    as the graph looks like |dw:1329981639188:dw|

    • 2 years ago
  34. cuddlepony
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    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
  35. aroberts47
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    oh it makes sense to me now

    • 2 years ago
  36. cuddlepony
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    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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