Derivation of the Chebyshev distance from the Minkowski distance

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I know that the Minkowski distance is defined as follows:

$$d(x,y)=\lim_{r\to\infty}\left(\sum_{k=1}^n|x_k-y_k|^r\right)^{1/r}$$

and I also read that the Chebyshev distance could be considered as a form of the Minkowski distance. I wanted to get a grasp by looking at the formula:

$$d(x,y)=\lim_{r\to\infty}\left(\sum_{k=1}^n|x_k-y_k|^r\right)^{1/r}=\underset{f}{\max}|x_{if}-x_{jf}|$$ but I cannot get the relationship. What I see is that if $r$ goes to $\infty$, then $\frac1r$ would be closer to $0$, so the (exp) would be closer to $1$, regardless if the absolute difference takes a big value from being raised to the power of $r$.

Anybody could help me to derive this equation?

Thanks