Evaluating a 0/0 limit with L'Hopital's Rule

Substituting $x=1$ gives numerator $1^1 - 1 = 0$ and denominator $1-1=0$, so the limit has the indeterminate form $0/0$, and L'Hopital's Rule applies. Write $x^x = e^{x\ln x}$ so that $\frac{d}{dx}(x^x) = \frac{d}{dx}e^{x\ln x} = e^{x\ln x}\cdot\frac{d}{dx}(x\ln x) = x^x(\ln x + 1)$, since $\frac{d}{dx}(x\ln x) = \ln x + 1$.

The derivative of the denominator $x-1$ is simply $1$. By L'Hopital's Rule, $\lim_{x\to 1}\frac{x^x-1}{x-1} = \lim_{x\to 1}\frac{x^x(\ln x + 1)}{1}.$ Substituting $x=1$: $1^1(\ln 1 + 1) = 1\cdot(0+1) = 1$. So the limit equals $1$.