To prove the limit exists using the fundamental definition. Here is how you proceed.

We must show that for everyϵ>0 there is δ>0 such that
if 0<|x−1|<δ, then |x2−1|<ϵ .
Finding δ is most easily accomplished by working backward. Manipulate the second inequality until it contains a term of the form x−1 as in the first inequality. This is easy here. First

|x2−1|=|x+1||x−1|.
In the above, there is unwanted factor of |x+1| , that must be bounded. If we make certain that δ<1
|x−1|<δ<1,
then
|x−1|<δ⟹|x−1|<1⟹−1<x−1<1
Adding 2 to the last inequality gives
1<x+1<3⟹|x+1|<3.
So, if
|x2−1|=|x+1||x−1|<3|x−1|<ϵ⟹|x−1|<ϵ3.
Now, select δ=min{1,ϵ3} .
Check: given ϵ>0 , let δ=min{1,ϵ3} . Then 0<|x−1|<δ implies that

|x2−1|=|x+1||x−1|<3|x−1|<3δ=3ϵ3=ϵ.

We must show that for every

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