# limit of $\frac{x^x -1}{x}$ as x tends to zero

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:

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

• 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. – André Nicolas Sep 26 '15 at 18:18
• @AndréNicolas thank you – user2804865 Sep 26 '15 at 19:04
• You are welcome. The basic L'Hospital's Rule strategy was fine. – 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.

• Why does it not exist? How would you show one side goes to something else. – Ahmed S. Attaalla Sep 26 '15 at 18:22
• @AhmedS.Attaalla Because $-\infty$ wouldn't be a limit per se. – Aleksandar Sep 26 '15 at 18:24
• I thought DNE was only used to show that the RH limit does not agree with the LH limit. – Ahmed S. Attaalla Sep 26 '15 at 18:26
• @AhmedS.Attaalla Not necessarily. – Aleksandar Sep 26 '15 at 18:28
• – Aleksandar Sep 26 '15 at 18:29

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: $$\lim_{x\to 0^+}\frac{x^x-1}{x}=\lim_{x\to 0^+}\frac{e^{x\log x}-1}{x\log x}\cdot\log(x) = \lim_{x\to 0^+}\log(x) = -\infty.$$

• Nice. Does it make a difference to use ln or log in this case? – GnP Sep 26 '15 at 18:32
• @gnp: no, it doesn't. For my standards, $\log$ means the natural logarithm as well as $\ln$. – 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 $$\frac{x^x-1}x=\frac{x\ln x+o(x\ln x)}x=\ln x+o(\ln x)\sim\ln x\xrightarrow[x\to 0^+]{}-\infty.$$