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Given the limit of [tex]\frac{x^2+2x}{x^2-3x}[/tex] as x approaches 0 equals [tex]\frac{-2}{3}[/tex] and that ε = .01, find the greatest c such that every δ between zero and c is good. Give an exact answer.

0 < |x-0| < δ

0 < |[tex]\frac{x^2+2x}{|x^2-3x}[/tex] + [tex]\frac{2}{3|}[/tex]| < ε

|[tex]\frac{x(x+2)}{|x(x-3)}[/tex] + [tex]\frac{2}{3|}[/tex]| = .01

|[tex]\frac{x+2}{|x-3|}[/tex]| = [tex]\frac{-197}{300}[/tex]

300(x+2) = -197(x-3)

300x + 600 = -197x + 591

497x = -9

x = [tex]\frac{-9}{497}[/tex]

But, my book got [tex]\frac{9}{503}[/tex]

0 < |x-0| < δ

0 < |[tex]\frac{x^2+2x}{|x^2-3x}[/tex] + [tex]\frac{2}{3|}[/tex]| < ε

|[tex]\frac{x(x+2)}{|x(x-3)}[/tex] + [tex]\frac{2}{3|}[/tex]| = .01

|[tex]\frac{x+2}{|x-3|}[/tex]| = [tex]\frac{-197}{300}[/tex]

300(x+2) = -197(x-3)

300x + 600 = -197x + 591

497x = -9

x = [tex]\frac{-9}{497}[/tex]

But, my book got [tex]\frac{9}{503}[/tex]

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