Answer
$$\lim_{x\to0^+}\frac{x^x-1}{\ln x+x-1}=0$$
Work Step by Step
$$A=\lim_{x\to0^+}\frac{x^x-1}{\ln x+x-1}$$
1) The problem in this exercise is to calculate $\lim_{x\to0^+}(x^x)$. Let's call this B.
$$B=\lim_{x\to0^+}(x^x)$$
In order to deal with $B$, we need to do a little transformation first.
Take $x^x=(e^{\ln x})^x=e^{x\ln x}$ (for $e^{\ln x}=x$)
So we have, $$B=\lim_{x\to0^+}(e^{x\ln x})$$
$$B=e^{\lim_{x\to0^+}(x\ln x)}$$
We now call $$C=\lim_{x\to0^+}x\ln x$$
Divide both numerator and denominator by $x$: $$C=\lim_{x\to0^+}\frac{\ln x}{\frac{1}{x}}$$
As $x\to0^+$, $(\ln x)\to-\infty$ and $\frac{1}{x}\to\infty$. An indeterminate form of $-\infty/\infty$, so we apply L'Hospital's Rule:
$$C=\lim_{x\to0^+}\frac{\frac{1}{x}}{-\frac{1}{x^2}}$$
$$C=\lim_{x\to0^+}-\frac{x^2}{x}$$
$$C=\lim_{x\to0^+}(-x)$$
$$C=-0=0$$
Consequently, $$B=e^{\lim_{x\to0^+}(x\ln x)}=e^0=1$$
That means, $$\lim_{x\to0^+}(x^x)=1$$
We now come back to the main question of $A$.
2) $\lim_{x\to0^+}(x^x-1)=1-1=0$ and $\lim_{x\to0^+}(\ln x+x-1)=-\infty$ (As $x\to0^+$, $\ln x+x-1\to(-\infty+0-1)=-\infty$)
That takes us to the form of $\frac{0}{\infty}$, which is no different from $0$.
Therefore, $$A=0$$
