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Limits of Arguments of Functions

  1. Nov 5, 2009 #1
    Hello,

    I am learning Calculus I and doing fine so far. I arrived at a spot that I can intuitively understand, but would like a more formal mathematical understanding of.

    It concerns limits of the arguments of functions.

    Consider:

    [tex]\displaystyle\lim_{x\to-\infty}ln(\frac{1}{x^2-4x})[/tex]

    Now, intuitively I can tell that the argument will go to 0 from the right, and that thus the natural log will never have a negative argument. The argument will approach 0, meanings the natural log will approach [tex]-\infty[/tex]

    All that is fine, but my problem is that after I show that:
    [tex]\displaystyle\lim_{x\to-\infty}\frac{1}{x^2-4x}=0[/tex]

    I don't know how to mathematically (rather than reasoning through it) show the next step, the one before this final one:

    [tex]\displaystyle\lim_{x\to-\infty}ln(\frac{1}{x^2-4x})=-\infty[/tex]

    Any help? I might guess that maybe the step would look like this:

    [tex]\displaystyle\lim_{\alpha\to0+}ln(\alpha)=-\infty[/tex]

    In this (almost surely wrong) attempt at creating an intermediate step, I take the fact that the argument approaches 0 from the right and call the argument [tex]\alpha[/tex]. This transforms the original problem, but yields the correct answer. Anyways, this was just to show my thoughts. What is the correct intermediate step?

    Thanks!
     
  2. jcsd
  3. Nov 5, 2009 #2

    dextercioby

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    I can give you a hint: the natural logarithm is a strictly increasing function on its domain of definition. This is a key point in showing that

    [tex] \lim_{x} \ln f(x) = \ln \left(\lim_{x}f(x)\right) [/tex].

    The same is valid for the inverse of the logarithm, namely the exponential function.
     
  4. Nov 5, 2009 #3
    Your post implies that I would simply write [tex]ln(0)[/tex] once I know the limit of the argument is 0. I was taught that [tex]ln(0)[/tex] is undefined, and not [tex]-\infty[/tex]. Was I taught incorrectly?
     
  5. Nov 5, 2009 #4

    lurflurf

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    It is a matter of semantics. It would be more correct to write
    log(0+)=-∞
    even then there remains the question is that the value of the function in the extended reals or mearly an indication of a type of divergence. ∞ is after all not a real number.
     
  6. Nov 5, 2009 #5
    I am very surprised by two things:

    1) That whether the value of [tex]log(0)[/tex] is [tex]-\infty[/tex] or undefined is a matter of semantics.
    2) That you can use the + notation inside the argument of a log, without dealing with limits (as you seem to have done).

    Are both points true?
     
  7. Nov 5, 2009 #6

    HallsofIvy

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    Saying that a limit is [itex]\infty[/itex] or [itex]-\infty[/itex] is just that it does NOT have a limit, in a particular way. Saying that [tex]ln(0)[/itex] is undefined, or that it is [itex]-\infty[/itex] are essentially the same thing.
     
  8. Nov 6, 2009 #7
    So, just to make sure I understand this correctly, it's accepted mathematically to write [tex]ln(0)=-\infty[/tex] as a result of bigubau's rule in order to arrive at the final answer of the limit in the original problem? If it is, great thanks! If it isn't, what is the mathematically accepted intermediate step between showing the argument's limit and showing the whole function's limit?

    Thanks again for all your help!
     
  9. Nov 9, 2009 #8
    From what I have understood of the responses to this thread, I could solve this problem the following way:

    [tex]\displaystyle\lim_{x\to-\infty}ln\left(\frac{1}{x^2-4x}\right)[/tex]

    [tex]\lim_{x} \ln f(x) = \ln \left(\lim_{x}f(x)\right) [/tex]

    [tex]\displaystyle\lim_{x\to-\infty}\frac{1}{x^2-4x}=0[/tex]

    [tex]\ln \left(0\right)=-\infty[/tex]

    Is this right?
     
  10. Nov 9, 2009 #9
    I think I remember doing something like using a substitution for problems like this before:
    [tex]R = \lim_{x\rightarrow -\infty}\frac{1}{x^2 - 4x} = 0[/tex]

    [tex]\lim_{R\rightarrow 0+}\ln(R) = -\infty[/tex]
     
  11. Nov 9, 2009 #10
    Wow! You mean to say that my guess for an intermediate step in my very first post was correct? Can anyone corroborate? Note that this is distinct from bigubau's suggestion as bigubau eventually calls for the natural log of the argument's limit, while Bohrok and I call for the limit of the natural log of the argument's limit.
     
  12. Nov 9, 2009 #11
    Basically the only difference was that bigabu was taking the natural log of 0 while you/I evaluated with that limit instead. The reason for doing the limit is so you don't do anything "illegal" like evaluating ln(0), or ln(0+), which I feel are illegal. Similar to saying 1/(0)2 = ∞ or 1/∞ = 0.
     
  13. Nov 9, 2009 #12
    You're dealing with limits, which do not necessarily have to do with what the function equals at a certain point, but instead focus on what the function is doing around that point.
     
  14. Nov 10, 2009 #13
    I have a feeling you read the title, saw someone was having trouble with limits, quickly scanned the thread, and wrote a standard statement about limits.
     
  15. Nov 12, 2009 #14
    No, I was merely agreeing with Bohrok's last statement.
     
  16. Nov 13, 2009 #15

    HallsofIvy

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    If f(x) is continuous at c and [itex]\lim_{x\to a} g(x)= c[/itex], then [itex]\lim_{x\to a}f(g(x))= f(\lim_{x\to a} g(x))[/itex]. Is that what you are asking about? That is pretty much the definition of "limit".
     
  17. Nov 13, 2009 #16
    the step is not rigorously correct because the outer function must be continuous at the limit of the inner function.
    You can instead use the logarithm identity: [tex] ln(\frac{1}{x^{2}-4x})=ln(1)-ln(x^{2}-4x)=-ln(x^{2}-4x)\stackrel{x\rightarrow -\infty}{\rightarrow}-\infty[/tex]
     
  18. Nov 13, 2009 #17
    Now I am very confused. Is natural log continuous at 0?

    If it isn't, then why would you write this in this thread? The reason I posted this problem was because ln(x) not being continuous at 0 meant that I could not use the rule you posted.
     
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