Let $(X,\mathcal{B},\mu)$ be a finite measure space. For $f\in L^{\infty}(X)$ with $\|f\|_{\infty}>0$, I want to prove that $\|f\|_{\infty}=\lim\limits_{p\rightarrow\infty}\frac{\|f\|_{p+1}^{p+1}}{\|f\|_p^p}$.
Here is my initial attempt. On one hand, $$\frac{\|f\|_{p+1}^{p+1}}{\|f\|_p^p}=\frac{\int_{X}|f|^{p+1}\,\mathrm{d}\mu}{\int_{X}|f|^{p}\,\mathrm{d}\mu}\leqslant\frac{\sup\limits_{x\in X}|f(x)|\int_{X}|f|^{p}\,\mathrm{d}\mu}{\int_{X}|f|^{p}\,\mathrm{d}\mu}=\sup\limits_{x\in X}|f(x)|=\|f\|_{\infty}.$$ For another direction, I take any $0<\varepsilon<\|f\|_\infty$ and define $X_\varepsilon\colon\!=\left\{x\in X\colon\|f\|_\infty\leqslant |f(x)|+\varepsilon\right\}$, then $$\frac{\|f\|_{p+1}^{p+1}}{\|f\|_p^p}=\frac{\int_{X}|f|^{p+1}\,\mathrm{d}\mu}{\int_{X}|f|^{p}\,\mathrm{d}\mu}\geqslant \frac{\left(\|f\|_{\infty}-\varepsilon\right)\int_{X_\varepsilon}|f|^{p}\,\mathrm{d}\mu+\int_{X\backslash X_\varepsilon}|f|^{p+1}\,\mathrm{d}\mu}{\int_{X}|f|^{p}\,\mathrm{d}\mu}.$$ Then I am stuck at this step. I speculate that the following equation holds: $$\lim\limits_{p\rightarrow\infty}\frac{\int_{X_\varepsilon}|f|^{p}\,\mathrm{d}\mu}{\int_{X}|f|^{p}\,\mathrm{d}\mu}=\frac{\mu(X_\varepsilon)}{\mu(X)}.$$ But I am not sure how to prove it. I tried using simple functions, such as $\varphi=\sum_{k=1}^{n}\alpha_k\chi_{X_k}$, but I failed. So, is this equation correct? If so, then how can I prove it? If not, then how can I prove the original problem?