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

$x^x$ does not approach $+\infty$ as $x \to 0^+$.

The limit $\lim_{x \to 0+} x^x$ is in fact equal to $1$, as we see by expressing $x^x = e^{x\ln{x}}$.

Thus, your limit $$\lim_{x \to 0+} \frac{1}{(\ln(x) + 1)x^x}$$ is in fact equal to $$\lim_{x \to 0+} \frac{1}{\ln(x) + 1} \cdot \lim_{x \to 0+} \frac{1}{x^x},$$ which is equal to $0 \cdot 1 = 0$.