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Simple limit problem (Which way is the best to solve?)

  1. Sep 14, 2007 #1
    I'm wondering what the best way is to solve:

    [tex]\lim_{x \rightarrow 3^{-}} \frac{x}{\sqrt{x^2-9}}[/tex]

    I'm pretty sure that f(x) is not equal to zero but I can't seem to manipulate it to cancel out (x-3). Also, when solving these types of problems, can you use the same rules as regular limits and substiute accordingly to obtain the limit as x approaches 3 from the right or left? Do people usually solve these problems analytically or graphically? I guess if you understand the function and can visualize it and know the laws and whatnot you can solve it analyitically right?

    In a nutshell, what is the simplest method into solving these problems accurately and quickly? (One that provides enough proof)
     
  2. jcsd
  3. Sep 14, 2007 #2
    What I'm thinking now is that since it is the limit as x approaches 3 from the left, you can just solve for the limit. However, if the limit does not exist, then you should revert back to the continuity rules or graph the function to solve it?
     
  4. Sep 14, 2007 #3

    Dick

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    Ask yourself first what do the numerator and denominator approach separately? This may be less of a problem than you think.
     
  5. Sep 14, 2007 #4

    Dick

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    If the limit doesn't exist, as I think you know, then it doesn't exist. You can't 'solve it'.
     
  6. Sep 14, 2007 #5
    Well the numerator approaches 3 from the left and the denominator approaches 0 from the left so would it be -3? How are you suppose to support your answer though? I'm learning continuity right now so am I suppose to use those rules?
     
  7. Sep 14, 2007 #6
    Best way would be to try and "pull out" the x^2 inside of the square root and try to cancel that x on the numerator.
     
  8. Sep 14, 2007 #7
    my algebra isn't that good, how would you do that? i saw a crazy one today ... i'm still trying to figure it out :(
     
  9. Sep 14, 2007 #8

    Dick

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    Pull out your calculator and start picking numbers close to 3 to approximate the limit (aside from the fact that in the left limit the denominator is not really well defined) and tell me how we should assign the limit? If the numerator has a well defined non-zero limit and the denominator doesn't there is no limit. I really don't think this calls for a 'proof'. As I said the denominator isn't even well defined. Proof enough for me.
     
  10. Sep 14, 2007 #9

    Dick

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    Since when is 3/0 anything close to -3?
     
  11. Sep 15, 2007 #10
    Oh, ok so it approaches -infinity right? I'm assuming you can't solve this analytically? Yeah bob, how can you pull the x^2 out of the sqrt?
     
  12. Sep 15, 2007 #11
    [tex](ax^b + c)^\frac{d}{e} = x^\frac{bd}{e}(a+\frac{c}{x^b})^\frac{d}{e}[/tex]

    @ILovePhysics hm..Is the denominator even defined on the left side? I graph it and it exists but can't find a formula equivilant o_O

    and yep you're correct -infinity is what it goes too, somehow o.o.
     
  13. Sep 15, 2007 #12
    Well, I just thought there was an asymptote at -3 but my brain wasn't functioning well and I just picked -1 for the denominator lol.
     
  14. Sep 15, 2007 #13

    Dick

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    1/x can be said to approach -infinity as x->0^- in some sort of reasonable sense. It's also reasonable to say the limit doesn't exist. This is worse. What is sqrt(-0.0001)?
     
  15. Sep 15, 2007 #14

    Dick

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    It is not defined, on that you are correct. Tough to get a limit then.
     
  16. Sep 15, 2007 #15
    Oh yeah, I forgot that infinities are never limits; is it because they are "infinity" and have no limit? It seems intuitive to me now. Thanks for your help guys!
     
  17. Sep 15, 2007 #16

    Dick

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    Right. It's because the answer to a limit question is generally supposed to be a real number. Infinities aren't real numbers. They definitely don't work in an epsilon-delta sense.
     
  18. Sep 15, 2007 #17

    dynamicsolo

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    The interval -3 < x < 3 isn't in the domain of the function, so this is worse than just "limit does not exist" -- the expression is just meaningless. I suspect this was a trick question given to the students to see if they are watching out for the behavior of the function.
     
  19. Sep 15, 2007 #18

    HallsofIvy

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    It's because "infinity" is not a real (or complex) number. Limits are always numbers. Saying "the limit is infinity" is just saying the limit does not exist for a specific reason.
     
  20. Sep 15, 2007 #19

    dynamicsolo

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    Those last two statements aren't quite right.

    An example of a limit that is *not* infinity, but rather does not exist is lim x-> infinity of sin x. Sine is a bounded function, but sin x does not approach any particular value between -1 and 1 more closely as x grows without limit.

    Another example is lim x-> 0 of ln x. The limit as x approaches 0 "from above" is minus infinity, but values x <= 0 are not in the domain of ln x, so the limit from below does not exist because the function has no definition there. So the "two-sided" limit does not exist because there is no way to define the limit (or the function) for x<0.

    Even for a function where the limit at a "from above" differs from the limit at a "from below" , the "two-sided" limit at a would be said not to exist. The individual "one-sided" limits can still each be finite and this would still be the case.

    What you are describing is referred to as an "infinite limit". An example would be
    lim x-> 0 of (1/(x^2)). Each of the "one-sided" limits at x = 0 is positive infinity, so the "two-sided" limit is said to be positive infinity. The limit *does* exist, but it is not a number. [The same cannot be said for lim x-> 0 of (1/x), because the "one-sided" limits are *differently-signed* infinities. So the "two-sided" limit at x=0 in this case does not exist.]

    "Infinity" is not used simply as a place-holder to signify the failure of a function to have a limit. There is a formal definition for an infinite limit which does place certain requirements on the behavior of the function.
     
    Last edited: Sep 16, 2007
  21. Sep 16, 2007 #20

    Dick

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    Thanks. Most illuminating.
     
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