For part A, (described here: http://www.cramster.com/solution/solution/1157440) I don't understand why they say |x-2| < 1 and why [itex]\delta[/itex] = min{1,ε/5}(adsbygoogle = window.adsbygoogle || []).push({});

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lim x^{2}+2x-5 = 3, x [itex]\rightarrow[/itex] 2

Let ε > 0 and L = 3.

|x^{2}+ 2x -5 -3| < ε

|x^{2}+ 2x - 8| < ε

|x+4||x-2| < ε

If |x-2| < 1, x [itex]\rightarrow[/itex] (1,3) [itex]\rightarrow[/itex] x+2 [itex]\in[/itex] (3,5)

|x-2| < 1 [itex]\rightarrow[/itex] |x+2| < 5

|x-2| < 1 [itex]\rightarrow[/itex] |x^{2}+2x - 8| = |x-4||x+2| < 5|x-4|

For ε > 0, 5|x-4| < ε [itex]\rightarrow[/itex] |x-4| < [itex]\frac{ε}{5}[/itex]

If [itex]\delta[/itex]= min{1,[itex]\frac{ε}{5}[/itex]}

|x-2| < [itex]\delta[/itex] [itex]\rightarrow[/itex] |x^{2}+ 2x - 8| < 5|x-4| < ε

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# Epsilon Delta Limit.

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