I was wondering if i could get any help with the following:

$$ \lim_{x \rightarrow 0^{+}} \frac{x^x -1}{x} $$.

thank you.

My attempt:

$$ \lim = \frac{e^{x \ln(x)} - 1}{x} = \lim ( x \ln (x)( x \ln(x) + 1)) $$

  • 1
    $\begingroup$ Your $\frac{x\ln x-1}{x}$ should have been $\frac{e^{x\ln x}-1}{x}$. And your differentiation of $x\ln x$ was not correct. $\endgroup$ – André Nicolas Sep 26 '15 at 18:18
  • $\begingroup$ @AndréNicolas thank you $\endgroup$ – user2804865 Sep 26 '15 at 19:04
  • $\begingroup$ You are welcome. The basic L'Hospital's Rule strategy was fine. $\endgroup$ – André Nicolas Sep 26 '15 at 19:06

Use L'Hopital's rule to get:

$$\lim_{x \to 0} x^x(\ln x+1)=\lim_{x \to 0}x^x \ln x-\lim_{x \to 0} x^x=(\lim_{x \to 0} x^x \ln x) -1=((\lim_{x \to 0} x^x)(\lim_{x \to 0} \ln x))-1=(1)(-\infty)-1=-\infty$$

The limit does not exist.


We may also avoid De l'Hopital theorem. Since $$ \lim_{t\to 0}\frac{e^t-1}{t}=1\quad\text{and}\quad \lim_{x\to 0^+} x\log(x)=0,$$ we have: $$ \lim_{x\to 0^+}\frac{x^x-1}{x}=\lim_{x\to 0^+}\frac{e^{x\log x}-1}{x\log x}\cdot\log(x) = \lim_{x\to 0^+}\log(x) = -\infty.$$

  • $\begingroup$ Nice. Does it make a difference to use ln or log in this case? $\endgroup$ – GnP Sep 26 '15 at 18:32
  • $\begingroup$ @gnp: no, it doesn't. For my standards, $\log$ means the natural logarithm as well as $\ln$. $\endgroup$ – Jack D'Aurizio Sep 26 '15 at 18:35

Use an asymptotic development:

$x^x=\mathrm e^{x\ln x}=1+x\ln x+o(x\ln x)$, hence $$\frac{x^x-1}x=\frac{x\ln x+o(x\ln x)}x=\ln x+o(\ln x)\sim\ln x\xrightarrow[x\to 0^+]{}-\infty.$$


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