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

Question

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:

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

Answer 1

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.

Answer 2

We may also avoid De L'Hospital 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: $$ \lim_{x\to 0^+}\frac{x^x-1}{x}=\lim_{x\to 0^+}\frac{e^{x\log x}-1}{x\log x}\cdot\log(x) = \lim_{x\to 0^+}\log(x) = -\infty.$$

Answer 3

Use an asymptotic expansion:

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