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Does anyone know the best way to show this limit does not exist?

  1. Jun 26, 2012 #1
    What's the most efficient way (other than just plotting it and eyeballing it) to show that

    [tex]
    \lim_{x\to 0} \frac{\ln (1-x) - \sin x}{1 - \cos^2 x}
    [/tex]

    does not exist?
     
  2. jcsd
  3. Jun 26, 2012 #2

    tiny-tim

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    l'Hôpital?
     
  4. Jun 26, 2012 #3


    Can't be. L'Hospital gives a sufficient condition for a limit to exist, not necessary, meaning: if the limit *after* doing L'H doesn't exist we cannot deduce the original limit doesn't exist either.

    DonAntonio
     
  5. Jun 26, 2012 #4

    micromass

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    Still, we can use L'Hopitals rule by noting it holds also for one-sided limits and for extended limits. So if we can show that the left-sides and right-sides limits of the derivatives are infinite, then so are the left-sided and rigt-sided limits of the original function.
     
  6. Jun 26, 2012 #5


    One way is using Taylor expansions: [tex]\log(1-x)=-(x+\frac{x^2}{2}+\frac{x^3}{3}+...)\,\,,\,\,\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\Longrightarrow \sin^2x=x^2-\frac{x^4}{3}+...[/tex] so[tex]\frac{\log(1-x)\sin x}{\sin^2x}=\frac{-2x-\frac{x^2}{2}-\frac{x^3}{6}-...}{x^2-\frac{x^4}{3}+...}=-\frac{1}{x}\frac{2+\frac{x}{2}+\frac{x^2}{3}+...}{1-\frac{x^2}{3}+...}[/tex]

    The second factor in the last expression above has limit [itex]\,2\,\text{when}\,x\to 0\,[/itex] , whereas the first factor has no limit, as it approaches [itex]\,-\infty\,\,\,or\,\,\,+\infty\,[/itex] according as whether [itex]\,x\to 0^+\,\,\,or\,\,\,x\to 0^-\,[/itex] , resp.

    DonAntonio
     
  7. Jun 26, 2012 #6
    [tex]\frac{\ln(1-x)-\sin x}{1-\cos^2x} = \frac{\ln(1-x)-\sin x}{\sin^2x} = \frac{\ln(1-x)}{\sin^2x} - \csc x[/tex]
    [tex]\lim_{x\to 0^-} \frac{\ln(1-x)}{\sin^2x} = +\infty[/tex] [tex]\lim_{x\to 0^+}\frac{\ln(1-x)}{\sin^2x} = -\infty[/tex][tex]\lim_{x\to 0^-}-\csc x = +\infty[/tex][tex]\lim_{x\to 0^+}-\csc x = -\infty[/tex]

    The left and right-sided limits aren't the same, so the limit doesn't exist.
     
  8. Jun 30, 2012 #7
    Does this mean that L'Hopital's rule is still good even if we rewrite the statement of the theorem in terms of left-hand (or right-hand) limits? (We obviously then just obtain a result about the left-hand or right-hand limit of the function.)
     
  9. Jun 30, 2012 #8
    You seem to be assuming that we can determine

    [tex]
    \lim_{x\to a} \Big( f(x) + g(x) \Big)
    [/tex]

    by considering

    [tex]
    \lim_{x\to a} f(x) + \lim_{x\to a} g(x),
    [/tex]

    even if the limits of [itex]f[/itex] and [itex]g[/itex] are [itex]\pm \infty[/itex]. Is this really the case?
     
  10. Jun 30, 2012 #9

    tiny-tim

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    yes … if we can use l'hôpital to prove that the left and right limits are ∞ and -∞, then we have proved that "the" limit does not exist :wink:
     
  11. Jun 30, 2012 #10
    A function has a limit at x0 iff its right and left limits at x0 exist and are equal.
     
  12. Jun 30, 2012 #11

    lurflurf

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    [tex]
    \lim_{x\to 0} x \frac{\ln (1-x) - \sin x}{1 - \cos^2 x}=-2
    [/tex]
     
  13. Jun 30, 2012 #12

    D H

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    Nope. That's [itex]\lim_{x\to 0} x \frac{\ln (1-x) - \sin x}{\sin x}[/itex]. It looks like you lost a power of 2 in the denominator.
     
  14. Jul 1, 2012 #13
    It depends on whether the infinite limits have the same sign. If, say, [itex]\lim_{x\to a} f(x) = \infty[/itex] while [itex]\lim_{x\to a} g(x) = -\infty[/itex], then we have to use other methods to investigate the limit. But, informally speaking, if [itex]\lim_{x\to a} f(x) = \lim_{x\to a} g(x) = \infty[/itex] (note: same sign), then close to a, f(x) is very large and g(x) is nonnegative, which means that f(x)+g(x)≥f(x)+0, which is still very large.
     
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