Limits using L'Hopital's rule $\lim_{x\to0^+} (x^{x^x-1})$

Solution 1:

Write everything in an accurate way: $$x^{x^x-1} = \exp ((\log x )(x^x-1)) = \exp \left( x\log^2 x\left( \frac{e^{x \log x}-1}{x \log x}\right) \right)$$ Now, since $\lim_{h \to 0}\frac{e^h-1}{h} =1$ and $\lim_{x \to 0}x \log^2 x = 0$ you get that the whole limit is $$\exp(0 \cdot 1) = e^0 = 1.$$

Solution 2:

HINT:

$$\lim_{x\to 0}\left(x^{x^x-1}\right)=\lim_{x\to 0}\exp\left(\ln\left(x^{x^x-1}\right)\right)=\lim_{x\to 0}\exp\left(\left(x^x-1\right)\ln(x)\right)=\exp\left(\lim_{x\to 0}\left(x^x-1\right)\ln(x)\right)$$

Solution 3:

APPROACH 1:

Although the OP requested an approach using L'Hospitals' Rule, I thought it might be instructive to present an approach that uses asymptotic analysis. To that end, we write for $x\to 0^+$

$$\begin{align} x^{x^x-1}&=e^{\log(x)\,\left(x^x-1\right)}=e^{\log(x)\,\left(e^{\log(x)\,x}-1\right)}\\\\ &= e^{x\log^2(x)+O\left(x^2\log^3(x)\right)} \end{align}$$

Inasmuch as both $\lim_{x\to 0^+}x\log^2(x)=0$ and $\lim_{x\to 0^+}x^2\log^3(x)=0$, we have

$$\lim_{x\to 0^+}x^{x^x-1}=1$$

APPROACH 2:

If one insists on using L'Hospital's Rule, we can proceed as follows.

$$\begin{align} \lim_{x\to 0^+}\log(x)\,\left(e^{\log(x)\,x}-1\right)&=\lim_{x\to 0^+}\frac{\left(e^{\log(x)\,x}-1\right)}{1/\log(x)}\\\\ &=\lim_{x\to 0^+}\frac{(1+\log(x))\,e^{x\log(x)}}{-\frac{1}{x\log^2(x)}}\\\\ &=-\lim_{x\to 0^+}x\log^2(x)(1+\log(x))\,e^{x\log(x)}\\\\ &=0 \end{align}$$

Therefore, we recover the aforementioned result that used asymptotic analysis

$$\lim_{x\to 0^+}x^{x^x-1}=1$$

as expected!