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Can someone help me out with $$\lim\limits_{x\rightarrow 0^+}x\log(1+x^{-1})?$$ I tried Taylor's expansion to no avail.

Prabath Hewasundara
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Teodorism
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    Hint: for large $x^{-1}$ the 1 becomes negligible. Eliminating it, you return to a familiar problem... – Ian Feb 18 '20 at 17:24
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    $x \rightarrow 0^+$? – Peter Szilas Feb 18 '20 at 17:25
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    @Ian, yes, you are right. Not sure how that makes the problem simpler. What is the familiar problem you are hinting at? – mjw Feb 18 '20 at 17:31
  • No. $\log$ is undefined for negative values. – Teodorism Feb 18 '20 at 17:54
  • @mjw I was thinking of pulling out the exponent and then taking the limit of just $x\log(x)$. Changing variables works too, of course. – Ian Feb 18 '20 at 17:55
  • That's why I asked below if L'Hopital is still applicable. Because it usually applies to functions that are differentiable in the open interval around the limit point. – Teodorism Feb 18 '20 at 17:58
  • @Ian, yes! That does come up all the time ($\lim x \log x $). Thanks! – mjw Feb 18 '20 at 17:58
  • Well, if a function only has a limit from the right, we can use L'Hopital's rule from the right. If we look at the derivation of L'Hopital's rule, it should be clear that we can limit it to one side. – mjw Feb 18 '20 at 18:01
  • mjw. Has been discussed already. $\log a$ is defined for $a>0$. $x \rightarrow 0^-$ would lead to a negative argument in $\log (1+1/x).$ – Peter Szilas Feb 18 '20 at 18:15
  • @PeterSzilas, agreed. Was discussed. Thanks. – mjw Feb 18 '20 at 19:11

4 Answers4

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$$\lim_{x\downarrow 0}x\log(1+x^{-1}) = \lim_{x\rightarrow\infty} \frac{\log(1+x)}{x}=0$$

by L'Hopital's rule.

mjw
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  • Does L'Hopital apply to limits of the form $ x\rightarrow 0^+$? (Because the limit in my question is from the right). – Teodorism Feb 18 '20 at 17:39
  • left and right limits at 0 are not equal for this function. Use the right-side limit. – Prabath Hewasundara Feb 18 '20 at 17:41
  • Yes. Here is a reference: http://mathworld.wolfram.com/LHospitalsRule.html – mjw Feb 18 '20 at 17:41
  • Is that why you changed the limit to $x\rightarrow \infty$ so that the numerator and denominator become differentiable in any open interval as $x$ goes to infinity? – Teodorism Feb 18 '20 at 17:42
  • No, actually, we could have applied L'Hopital's rule to $\lim \frac{\log(1+\frac{1}{x})}{\frac{1}{x}}.$ I made the change-of-variables $w=1/x$ because the derivative is a little easier in both numerator and denominator. – mjw Feb 18 '20 at 17:44
  • Okay, using notation to indicate we are taking the limit from the right. Thanks for pointing out this subtlety. – mjw Feb 18 '20 at 18:09
  • mjw. Well done. – Peter Szilas Feb 18 '20 at 21:24
  • @PeterSzilas, Thank you. – mjw Feb 20 '20 at 01:30
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$$\lim_{x\rightarrow 0}x\log(1+x^{-1})=\lim_{x\rightarrow 0}x\log(x^{-1}(x+1))=\lim_{x\rightarrow 0}x\log(x+1)-\lim_{x\rightarrow 0}x\log x = 0-0=0$$

We know that $\lim_{x\rightarrow 0}x\log x=\lim_{x\rightarrow 0}\frac{\log x}{\frac{1}{x}} $ using L'Hopital's rule

$$\lim_{x\rightarrow 0}x\log x = \lim_{x\rightarrow 0}\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim_{x\rightarrow 0}x=0$$

K.K.McDonald
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$$\lim_{x\rightarrow 0}\ln\left(1+\frac{1}{x}\right)^{x}=\lim_{x\rightarrow 0}\frac{\ln\left(1+\frac{1}{x}\right)}{\frac{1}{x}}=\frac{\frac{-\frac{1}{x^{2}}}{1+\frac{1}{x}}}{-\frac{1}{x^{2}}}=\frac{x+1}{x}=1$$

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By $1+\frac1x=y \to \infty$ we have

$$x\log(1+x^{-1})=\frac1{y-1}\cdot \log y=\frac y{y-1}\cdot \frac{\log y}y \to 1\cdot 0 =0$$

using that eventually $\log y\le \sqrt y$.

user
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