How to show that f(x)=x^2 is continuous at x=1?

To prove the limit exists using the fundamental definition. Here is how you proceed.
We must show that for every ϵ>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.