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Limit and continuity

  1. Aug 20, 2008 #1
    f(x) = sqrt(x+3), x > -3
    3 x2-5, x < -3


    first question would be limit x -> -3-, limit x -> -3+, limit x -> -3
    the answer would be [-1, 1, undefined]

    But, i only got the 3rd answer correct, the first two are wrong?? but i don't get it why, the previous similar question, i got it correct, but here i got it wrong??
     
  2. jcsd
  3. Aug 20, 2008 #2
    When you are being asked for [tex] \lim_{x \rightarrow -3^{-} [/tex] that means you are approaching -3 from the left side so look at which part of your f deals with the values "to the left" of -3 i.e. -4, -3.5, -3.1, -3.01, ...., -3.0000001, ... That part should be [tex] 3x^{2} - 5 [/tex] and the values I'm getting are 43, 31.75, 23.83, 22. Similarly for [tex] \lim_{x \rightarrow -3^{+} [/tex] you deal with values "to the right" of -3 which you would use [tex] \sqrt{x+3} [/tex] so once again pick values like -2, -2.5, -2.9, ..., -2.99999, -2.99999999999999 you should get 1, .707107, .003162, 0.0000001. Does that make sense? The reasons why you got the 3rd part correct is because for the limit to exist, you need to have the limit be the same regardless of which side you approach from which is obviously not the case here.
     
  4. Aug 20, 2008 #3

    tiny-tim

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    Hi fr33pl4gu3! :smile:
    Why did you think lim sqrt(x+3) as x -> -3+ was 1? :confused:

    If we know that, then we can help you. :smile:
     
  5. Aug 20, 2008 #4
    so the answer should be 43, 1, undefined, correct??
     
  6. Aug 20, 2008 #5

    tiny-tim

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    Are you using -4 instead of -3? :confused:

    Always show us your full calculation, not just your answer, if you want us to understand where you're going wrong. :smile:
     
  7. Aug 20, 2008 #6
    Does both of the final answer has to be equal??
     
  8. Aug 20, 2008 #7
    How do you get 43? As I said, for the limit to exist, you must get the same answer if you approach from the right hand side and left hand side.
     
  9. Aug 20, 2008 #8
    lhs = 3(-4)2-5=43;
    rhs = sqrt(1846+3)=43;

    correct??
     
  10. Aug 20, 2008 #9
    Did you read my post where I pretty much solved the problem for you?
     
  11. Aug 20, 2008 #10
    yes, i did try, but this question is a quiz system, so it say is incorrect, so i have no idea too, i use 43 just like you said, lhs = x = -4, rhs = x = 1846.
     
  12. Aug 20, 2008 #11
    Why are you all of a sudden using 4? Wasn't your original question to have x approaching 3 from both sides?
     
  13. Aug 20, 2008 #12
    Yep, thanks, i solve it, type wrongly, supposed to be typing the final answer but i always type the value of x, so sorry.
     
  14. Aug 21, 2008 #13

    HallsofIvy

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    Since 3x2- 5 is a continuous function, the limit, as x goes to -3 from below is just 3(-3)2- 5 which is definitely NOT -1.
    Since [itex]\sqrt{x+ 3}[/itex] is a continuous function for [itex]x\ge -3[/itex] The limit as x goes to -3 from above is just [itex]\sqrt{-3+3}[/itex] which is NOT 1.

    Since you have shown no work at all, I have no idea how you got those answers.

    (And since this has nothing at all to do with differential equations, I am moving it to the "Calculus and Beyond" homework section.)
     
  15. Aug 21, 2008 #14

    HallsofIvy

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    "Just like you said"? No one but you has mentioned "43"!
     
  16. Aug 21, 2008 #15

    tiny-tim

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    43

    Yes they did :smile:: ​
     
  17. Aug 21, 2008 #16
    Ohh I see how he got 43 now. Thanks tiny-tim. In my effort to show him what it means to approach from different sides, he took it to mean something completely different.
     
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