<s>[INST] Epsilon-delta limit definition I got $\lim_{x \rightarrow 3} \frac{x^2-2x+1}{x-2} = 4$
I need to prove that by delta epsilon. I came to delta ={1/2epsilon,1/2}
Is that right? If not can you explain me how?
 [/INST] |$\frac {(x-3)(x-3)}{x-2}|$< $\epsilon$
$\delta$ <= $\frac{1}{2}$
|x-3|<$\delta$ <=$\frac{1}{2}$
-$\frac{1}{2}$ < x-3 < $\frac{1}{2}$
$\frac{1}{2}$ < x-2 < $\frac{3}{2}$
2>$\frac {1}{x-2}$>$\frac{2}{3}$
$\delta$ = min{$\frac{1}{2}$$\epsilon$, $\frac{1}{2}$}
|$\frac {(x-3)(x-3)}{x-2}|$ < $\delta$*2 = $\epsilon$
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