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Find $\displaystyle\lim_{x\to0+}\frac{\ln(x^2)} x$

We have here a limit of the form: $"-\infty\cdot \infty"$ or $"\frac {-\infty}0"$.

LHR won't help with this limit since it diverges.

I tried to squeeze it but it didn't quite work...

Any hints please?

Another question about the use of LHR, if we get an expression of the form $\frac {-\infty}{\infty}$, are we allowed to use LHR? It's essentially the same as $\frac 0 0$ but from different sides.

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  • $\begingroup$ LHR also works when the limit of the quotients of the derivatives is $\infty$. $\endgroup$ – Carol Feb 9 '15 at 15:41
  • $\begingroup$ Just to be clear, when you say LHR, you're talking about L'Hôpital's Rule, correct? Yes, it is applicable for $\frac{\infty}{\infty}$ forms, but not for $\frac{\infty}{0}$. $\endgroup$ – Matthew Leingang Feb 9 '15 at 15:41
  • $\begingroup$ $-\infty\cdot\infty$ is not an indeterminate form like $0\cdot\infty$. It is always $-\infty$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Feb 9 '15 at 16:57
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use the substitution $x\to x^{-1}$, then we get $\lim _{x\to \infty}\frac {\ln x^{-2}}{x^{-1}}=\lim_{x\to \infty}-2x\ln(x)\to-\infty $

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  • $\begingroup$ When you use this substitution doesn't it flip the limit as well, so would be $1/-\infty$? $\endgroup$ – GinKin Feb 9 '15 at 15:58
  • $\begingroup$ the change goes from $x\to 0^+$ to $x\to \infty$ $\endgroup$ – Alan Feb 9 '15 at 16:29
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You are right that the form is $-\infty\cdot\infty$, but that just results in $-\infty$, so that is your limit. You don't need L'Hopital or anything else to go further.

Your second question: Yes. That is a standard addition in calculus texts about L'Hopital's rule.

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  • $\begingroup$ Why is it $-\infty$? $\endgroup$ – GinKin Feb 9 '15 at 15:50
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We use LHR for indeterminate forms like $\frac{\infty}{\infty}$. However, $-\infty\cdot\infty=-\infty$. If $\lim_{x\to 0^+}f(x)=-\infty$, and $\lim_{x\to 0^+}g(x)=\infty$, then $f(x)g(x)<0$ as $x\to 0+$, and $|f(x)g(x)|\to\infty$.

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  • $\begingroup$ Why is it $−∞$? $\endgroup$ – GinKin Feb 9 '15 at 15:50
  • $\begingroup$ Because as $x$ approaches $0$ from the right, $f(x)g(x)$ is negative. $\endgroup$ – Tim Raczkowski Feb 9 '15 at 15:52
  • $\begingroup$ So in the same way we have $-\infty \cdot -\infty=+\infty$? $\endgroup$ – GinKin Feb 9 '15 at 16:10
  • $\begingroup$ Yes, that would be correct. $\endgroup$ – Tim Raczkowski Feb 9 '15 at 16:11
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the limit does not exist beacause $\ln(0+) = -\infty$ and $\lim_{x \to 0+} \dfrac{1}{x} = \infty$ and $-\infty * \infty = -\infty.$

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  • $\begingroup$ The limit in the OP's question is $ x \to 0+$ $\endgroup$ – user159517 Feb 9 '15 at 15:47
  • $\begingroup$ thanks @user159517. i will fix it. $\endgroup$ – abel Feb 9 '15 at 15:54

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