극한의 엄밀한 정의 - 엡실론과 델타

\(\lim_{(x,y)\to(3,2)}\frac{y}{x-1}=1\) 의 증명

First we prove a set of inequalities.

Assume \(\sqrt{(x-3)^2+(y-2)^2}<\delta\).

(1) \(|x-3|\leq \sqrt{(x-3)^2+(y-2)^2}<\delta\) implies \(|x-3|<\delta\)

(2) \(|y-2|\leq \sqrt{(x-3)^2+(y-2)^2}<\delta\) implies \(|y-2|<\delta\)

(3) \(|y-x+1|=|(y-2)-(x-3)|\leq |y-2|+|x-3|<2\delta\)

(4) By rewriting \(|x-3|<\delta\) as \(|(x-1)-2|<\delta\), we get \(2-\delta<|x-1|<2+\delta\).

Let \(\epsilon>0\) be given. Let \(\delta\) be the minimum of \(\{1,{\epsilon}/2\}\).

Then from above inequalities, we can say

\(|y-x+1|<2\delta\leq \epsilon\) by (3)

\(|x-1|>2-\delta\geq 1\) (by (4)) so \(\frac{1}{|x-1|}<1\).

\(|\frac{y}{x-1}-1|=|\frac{y-x+1}{x-1}|<|y-x+1|<\epsilon\)

Therefore,

\(\lim_{(x,y)\to(3,2)}\frac{y}{x-1}=1\).

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