(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Prove that if [tex] |x-x_0| < \textrm{min} \bigg ( \frac{\epsilon}{2|y_0|+1},1 \bigg )[/tex] and [tex]|y-y_0| < \frac{\epsilon}{2|x_0|+1}[/tex] then [tex]xy-x_0y_0<\epsilon[/tex]

2. Relevant equations

We can use basic algebra and the following axioms:

For any number [tex]a[/tex], one and only one of the following holds:

(i) [tex]a=0[/tex]

(ii) [tex]a[/tex] is in the collection [tex]P[/tex]

(iii) [tex]-a[/tex] is in the collection [tex]P[/tex]

Note: A number [tex]n[/tex] is the collection [tex]P[/tex] if and only if [tex]n>0[/tex]

If [tex]a[/tex] and [tex]b[/tex] are in [tex]P[/tex], then [tex]a+b[/tex] is in [tex]P[/tex].

If [tex]a[/tex] and [tex]b[/tex] are in [tex]P[/tex], then [tex] a \cdot b [/tex] is in P.

We may also use the following consequences of the above axioms:

For any numbers [tex]a[/tex]and [tex]b[/tex], one and only one of the following holds:

(i) [tex]a=b[/tex]

(ii) [tex]a < b[/tex]

(iii)[tex] a > b[/tex]

For any numbers [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex], if [tex]a<b[/tex] and [tex]b<c[/tex], then [tex]a<c[/tex].

For any numbers [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex], if [tex]a<b[/tex], then [tex]a+c<b+c[/tex].

For any numbers [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex], if [tex]a<b[/tex] and [tex]0<c[/tex], then [tex]ac<bc[/tex].

3. The attempt at a solution

I'm not even sure where to start.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Proof of Inequality

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