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lim x^2 + x - 12 = 7

x-> 3 ------------

x-3

here is my solution so far but think it is wrong

Given E > 0 and D > 0 such that

|x^2 + x - 12 |

|------------ - 7| < E whenever 0 < |x-3| < D

| x-3 |

|(x-3)(x+4) |

|------------ - 7| < E whenever 0 < |x-3| < D

| x-3 |

|x + 4 - 7| < E

|x - 3| < E

Not sure where to go from here or if I am even close.

Thanks

Peter

PS I am new to proofs