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Prove the statement using the epsilon, delta definition of limit.

limx3x2+x12x3=7\lim _{x \rightarrow 3} \frac{x^2+x-12}{x-3}=7

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Let ε\varepsilon be a given positive number. We have a=3a=3 and L=7L=7, we need to find a number δ\delta such that

if   0<x3<δ   then   x2+x12x37<ε\text{if }~~ 0\lt |x-3|\lt\delta ~~\text{ then }~~ |\dfrac{x^2+x-12}{x-3}-7|\lt\varepsilon

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