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

  • $\begingroup$ Hint: for large $x^{-1}$ the 1 becomes negligible. Eliminating it, you return to a familiar problem... $\endgroup$ – Ian Feb 18 '20 at 17:24
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    $\begingroup$ $x \rightarrow 0^+$? $\endgroup$ – Peter Szilas Feb 18 '20 at 17:25
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    $\begingroup$ @Ian, yes, you are right. Not sure how that makes the problem simpler. What is the familiar problem you are hinting at? $\endgroup$ – mjw Feb 18 '20 at 17:31
  • $\begingroup$ No. $\log$ is undefined for negative values. $\endgroup$ – Teodorism Feb 18 '20 at 17:54
  • $\begingroup$ @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. $\endgroup$ – Ian Feb 18 '20 at 17:55

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

by L'Hopital's rule.

  • $\begingroup$ Does L'Hopital apply to limits of the form $ x\rightarrow 0^+$? (Because the limit in my question is from the right). $\endgroup$ – Teodorism Feb 18 '20 at 17:39
  • $\begingroup$ left and right limits at 0 are not equal for this function. Use the right-side limit. $\endgroup$ – Gune Feb 18 '20 at 17:41
  • $\begingroup$ Yes. Here is a reference: mathworld.wolfram.com/LHospitalsRule.html $\endgroup$ – mjw Feb 18 '20 at 17:41
  • $\begingroup$ 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? $\endgroup$ – Teodorism Feb 18 '20 at 17:42
  • $\begingroup$ 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. $\endgroup$ – mjw Feb 18 '20 at 17:44

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

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

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


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

  • $\begingroup$ Are you sure about this? $\endgroup$ – Gune Feb 18 '20 at 17:29

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