## 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<|x1|<δ, then |x21|<ϵ. Finding δ is most easily accomplished by working backward. Manipulate the second inequality until it contains a term of the form x1 as in the first inequality. This is easy here. First
|x21|=|x+1||x1|.
In the above, there is unwanted factor of |x+1|, that must be bounded. If we make certain that δ<1
|x1|<δ<1,
then
|x1|<δ|x1|<11<x1<1
Adding 2 to the last inequality gives
1<x+1<3|x+1|<3.
So, if
|x21|=|x+1||x1|<3|x1|<ϵ|x1|<ϵ3.
Now, select δ=min{1,ϵ3}. Check: given ϵ>0, let δ=min{1,ϵ3}. Then 0<|x1|<δ implies that

|x21|=|x+1||x1|<3|x1|<3δ=3ϵ3=ϵ.