When $|f|,\log|f| \in L^1$, we may prove this by recognizing the definition of the derivative:
Indeed, we can take logs to see that $\lim_{p \to 0} \frac{\log\int |f|^p d\mu}{p}=\frac{d}{dp} \int |f|^pd\mu \bigg|_{p=0} = \int \frac{d}{dp}|f|^p\bigg|_{p=0}d\mu = \int \log|f|d\mu.$
The reason we can put the derivative inside the integral sign is because we know that $\frac{d}{dp} |f|^p = |f|^p\log|f|$ which is bounded (uniformly in $p \leq 1/2$) by $2|f| + |\log|f||\in L^1$. Indeed, $|f|^p|\log|f|| \leq |\log|f||$ when $|f|\leq 1$, and $|f|^p |\log |f||\leq |f|^{1/2}|\log|f||\leq2|f|$ when $|f|\geq 1$ (since $|\log u| \leq 2u^{1/2}$ for $u \geq 1$). Thus applying Theorem 3.5.1 in these notes gives the second equality above.
Note that this is essentially the same proof given in the other answer above, the main point is to use dominated convergence. I just wanted to exposit it in a slightly different way.