MHB What is the limit of a circumference as the power approaches infinity?

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Show that $\displaystyle\lim_{p\to +\infty}C_p ((0,0); 1)=C_{\infty}((0,0);1).$

Hello, I think that this limit is infinite. At $C_p=\{(x,y)\in \mathbb{R}^2: |x|^p+|y|^p=1, \, p>1\}$, then is reasonably think. But how can show this?
 
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An expression equivalent is $\displaystyle\lim_{p\to +\infty} \left|x \right|^p+\left|y \right|^p=\max\{\left|x \right|, \left|y \right|\}.$

Can use the definition $\varepsilon-N$?
 
Julio said:
An expression equivalent is $\displaystyle\lim_{p\to +\infty} \left|x \right|^p+\left|y \right|^p=\max\{\left|x \right|, \left|y \right|\}.$
I don't think this is true. If $|x|<1$, then $\lim_{p\to\infty}|x|^p=0$.
 

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