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I'm having trouble with a step;I was wondering if someone could shed some light on me. Thanks.

Prove that lim x->-2 (x^2-1)=3

0<|f(x)-L|<epsilon whenever 0<|x-a|<delta

0<|(x^2-1)-3|<epsilon whenever 0<|x-(-2)<delta

= |x^2-4|<epsilon

= |(x-2)(x+2)|<epsilon

= |x-2||x+2|<epsilon

= If |x-2|<C (C=constant) then |x-2||x+2|<C|x+2|

= C|x+2|<epsilon = |x+2|<epsilon/C

Assume |x+2|<1 so -1<x+2<1 = -3<x<-1 = -5<x-2<-3

After this step -5<x-2<-3, I'm inclined to set |x-2|<-3, but book states that |x-2|<5. I'm lost at this step, should it be|-5|? Anyways, any explanations would be great.

Thanks,

Chris

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# Homework Help: Precise definition of a limit

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