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.