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Expressing limits mathematically

  1. Jul 19, 2009 #1

    Mentallic

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    I haven't had much exposure to limits and trying to understand it by myself has lead me into deep realms of frustration and heartache, so here I am to ask a simple questions (not homework).

    I have a decent understanding of asymptotes and it's the only reason that I can explain through logic that:

    [tex]\stackrel{lim}{x \rightarrow \infty}\sqrt{x^2+c}=x[/tex] for some constant c, since as x gets large, the constant loses its value and this tends towards [tex]\sqrt{x^2}=x[/tex]

    I want to know how to show this mathematically though. Oh and it would be nice if someone gives me the latex code to express limits properly :smile:
     
  2. jcsd
  3. Jul 19, 2009 #2
    You've already expressed it mathematically. You could factor out an x to make it more clear I suppose to get [itex]\sqrt{x^2+c}=x\sqrt{1+\frac{c}{x^2}}[/itex] and then in the limit as x-> inf [itex]\frac{c}{x^2}\rightarrow0[/itex] therefore you get x
     
  4. Jul 19, 2009 #3

    Office_Shredder

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    There are two common ways that functions 'approximate' each other near infinity. Using the context of your chosen functions

    1) [tex] \lim_{x \rightarrow \infty } \frac{ \sqrt{x^2+c} }{x} = 1 [/tex]

    so the ratio of the functions goes to 1. Or

    2) [tex] \lim_{x \rightarrow \infty} (\sqrt{x^2+c} - x) = 0[/tex]

    Number 2 is the harder criterion; if you can satisfy 2, you always satisfy 1. In most cases however, you're interested in ratios of functions near infinity, so 1 is probably used more often
     
  5. Jul 19, 2009 #4
    Whooaaaa. The limit has three options: a constant, +/- infinity, or doesn't exist. It can't equal an expression containing the variable that is what the limit is varying. You are right that the c doesn't affect the limit, but the limit doesn't equal x because x is approaching infinity. The answer is
    [tex]\lim_{x\to\infty} \sqrt{x^2+c} = \infty[/tex]

    (To see how to format in LaTeX, click the output to see the code.)
     
  6. Jul 20, 2009 #5
    I think he meant to say that [tex] \lim_{x \rightarrow \infty} \sqrt[]{x^2 +c} = \lim_{x \rightarrow \infty} x = \infty [/tex]
     
  7. Jul 20, 2009 #6
    Well the post I was responding to was deleted I guess, but the definition for a limit to be infinity as x approaches infinity is:
    For all M>0, there exists a number N>0 such that if x>N, then f(x)>M.
     
    Last edited: Jul 20, 2009
  8. Jul 20, 2009 #7

    Mentallic

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    I really like the way maverick has done it.

    Yeah sorry. I meant to say it is asymptotic to y=x. I was thinking that is how it's expressed. I guess I was wrong.
     
  9. Jul 20, 2009 #8
    Heh, sorry I caught the oversight. I knew something wasn't right when I was treating x as a variable and a number. Anyways, I worked out the proof under the correct definition and it is rather straightforward, though rather ugly, since you have to take care of the case when M is small and c is large and negative.
     
  10. Jul 20, 2009 #9
    Well in the example the conditioning is kinda tenous but [itex]\frac{c}{x^2}\rightarrow0[/itex] faster than x-> infinity.
     
  11. Jul 22, 2009 #10
    My guess is that the OP is asking for the rigorous proof. This is done using something called a delta-epsilon proof.
     
  12. Jul 22, 2009 #11

    Mentallic

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    What do you mean by a rigorous proof? Is something wrong with what has been posted thus far? And this delta-epsilon proof is new to me.
     
  13. Jul 22, 2009 #12
    I already posted the definition for this type of limit in post #6. Here's the proof.

    Let M>0. Let [tex]N=\sqrt{|M^2-c|}[/tex], and then N>0. If x>N, then
    [tex]
    \begin{align*}
    x &> \sqrt{|M^2-c|} \\
    x^2 &> |M^2-c| \\
    &> M^2-c \\
    x^2+c &> M^2 \\
    \sqrt{x^2+c} &> M
    \end{align*}
    [/tex]
    Thus, for any M>0, there exists an [tex]N=\sqrt{|M^2-c|}>0[/tex] such that x>N implies [tex]\sqrt{x^2+c}>M[/tex]. Therefore,
    [tex]
    \lim_{x\to\infty} \sqrt{x^2+c} = \infty
    [/tex]
     
  14. Jul 22, 2009 #13

    Mentallic

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    Sorry for my ignorance, but while I could somewhat follow the steps you made, I can't understand how you came to the final conclusion that the limit = infinite.
     
  15. Jul 22, 2009 #14
    Delta-Epsilon proofs are a pain in the neck and take a little work to understand - that is actually the reason I was so vague. To make it worse there are a lot of different ways people present the proofs.

    The basic idea is that as you approach the value at which your limit is being evaluated at, the value of the function has to get progressively closer to the limit. In the proofs, delta can be thought of as the change in x, and epsilon is the error of your limit.
     
  16. Jul 22, 2009 #15
    No problem. This is actually all overkill, as you are just learning limits and the techniques for evaluating them, but this is good information to know, as it is the theory behind limits.

    Definition:
    [tex]
    \lim_{x\to\infty} f(x) = \infty
    [/tex]
    if for all M>0, there exists a number N>0 such that if x>N, then f(x)>M.

    This means that no matter how big we choose M, we can find an x large enough so that f(x)>M. Thus as x gets larger, f(x) increases without bound, and the limit is infinity. Don't dwell too much on this if it doesn't click at first. Just get good at evaluating limits with sound techniques.
     
  17. Jul 22, 2009 #16

    Mentallic

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    It is pretty overwhelming...

    n!kofeyn, what if [tex]f(x)=\frac{1}{x}[/tex] ? The limit wont increase to infinite as x increases without bound. How is the proof applied for this function?
     
  18. Jul 22, 2009 #17
    Don't worry about epsilon-delta proofs of calculus. If you're using limits then for your course/application that's rigorous enough. Although for your own edification you might like to know that the limits approach of Newton was tossed out about 150 years ago due to it not being rigorous. For example, if you look at the limit proof of the power law there's actually a step where Newton says that 0^0=1.. Which is tenuous at best. Also the notion of dy and dx being "aribitrarily small quantities" is really very vauge. Regardless calculus isn't actually viewed that way nowaday. However, I assume for you the limit approach is just fine and, as I've showed, you get x, however since your limit is as x-> infinity then x=infinity. So the limit is really infinity. When I said it was x it was my bad. In physics we tend to think in the limit towards infinity as if we simply have "large values of x" in which case we get x to a good approximation (the larger the value of x the better the approximation) because 1/x^2 tends to zero much faster. However, in the true mathematical limit, x=infinity so we get infinity.
     
  19. Jul 22, 2009 #18
    There are different definitions for the different types of limits. There are limits with x approaching a finite number, and the limit is finite. There are limits, as above, with x approaching plus/minus infinity, and the limit is plus/minus infinity. There are limits approaching plus or minus infinity, and the limit is finite.

    The definition we would use for the problem you state here is:
    [tex]
    \lim_{x\to\infty} f(x) = L
    [/tex]
    if for all [itex]\varepsilon>0[/itex], there exists a number N>0 such that if x>N, then
    [tex]
    |f(x)-L|<\varepsilon .
    [/tex]

    Let's prove that
    [tex]
    \lim_{x\to\infty} \frac{1}{x} = 0
    [/tex]
    Note here that f(x)=1/x and L=0. Let [itex]\varepsilon>0[/itex]. By the Archimedean property, there exists a natural number N, such that [itex]N>\frac{1}{\varepsilon}[/itex]. Then [itex]\frac{1}{N}<\varepsilon[/itex]. If x>N, then
    [tex]
    \frac{1}{x} < \frac{1}{N} < \varepsilon .
    [/tex]
    Thus, for any [itex]\varepsilon>0[/itex], there exists a number N>0, such that [itex]|\frac{1}{x}-0|<\varepsilon[/itex]. Therefore,
    [tex]
    \lim_{x\to\infty} \frac{1}{x} = 0 .
    [/tex]

    Like I said, don't worry about these at this point, but it doesn't hurt to be introduced to the ideas. If you are interested in them, these are taught in some university calculus courses, like an honors course, and are definitely taught in an introductory analysis course.
     
  20. Jul 22, 2009 #19
    I understand what you're saying and getting at, but it's a bit misleading. These limits are perfectly rigorous. While the epsilon-delta proofs aren't taught at first, they are what back up all the limit properties used in beginning calculus classes. Newton's approach wasn't necessarily tossed out, rather it was simply reworked with the rigorous definitions of limits in terms of epsilon-deltas.
     
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