How do you prove that the $lim_(x to -2)(x^2 -5) = -1$ using the formal definition of a limit?

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Answer 1 · 2017-04-13T11:48:43

We want to show:

$forall epsilon>0$ , $exists delta>0$ s.t. $0<|x-(-2)|< delta Rightarrow |x^2-5-(-1)|< epsilon$

Please see the details below.

$forall epsilon>0$ , $exists delta=min{epsilon/5,1}>0$ s.t. $0<|x-(-2)|< delta$

$Rightarrow|x+2|< epsilon/5$

and

$Rightarrow|x+2|<1 Rightarrow-1 < x+2 < 1 Rightarrow-5< x-2 <-3 Rightarrow |x-2|<5$

So, we have

$|x^2-5-(-1)|=|x^2-4|=|x+2||x-2|< epsilon/5 cdot 5=epsilon$

Hence, $lim_(x to -2)(x^2-5)=-1$

I hope that this was clear.

How do you prove that the lim_(x to -2)(x^2 -5) = -1lim_(x to -2)(x^2 -5) = -1 using the formal definition of a limit?