Prove the limit (as x goes to 3) x^(2)=9
I expect Plato was extremely surprised that you could not use the fact that $\displaystyle \displaystyle \begin{align*} \lim_{x \to a}f(x) = f(a) \end{align*}$ if the function is continuous at $\displaystyle \displaystyle \begin{align*} x = a \end{align*}$. If you need an $\displaystyle \displaystyle \begin{align*} \epsilon - \delta \end{align*}$ proof, you need to STATE it.
The problem is that we cannot read your mind! You said "Prove that $\displaystyle \lim_{x\to 3} x^2= 9$" and there are many different ways to do that, depending on what you know. One method, if you know about continuous functions and, in particular, that all polynomials are continous for all x, then you have immediately that the limit is $\displaystyle 3^2= 9$, as Plato said. If that was not what you wanted, well, we have no way of knowing that unless you tell us exactly what you want. That is one reason why we repeatedly ask people to show what they have tried- so that we will at least know what basis they are starting from.
If you need to use an "epsilon-delta" proof then we needed to know that from the start. To have $\displaystyle |f(x)- L|= |x^2- 9|= |(x-3)(x+3)|< \epsilon$. That can be reduced to $\displaystyle |x-3||x+3|< \epsilon$ or $\displaystyle |x-3|< \frac{\epsilon}{|x+3|}$ so to be able to say $\displaystyle |x-3|< \delta$ implies that, you need a lower bound on $\displaystyle \frac{\epsilon}{|x+3|}$ and so an upper bound on |x+ 3|.
We are looking at values of x close to 3 so we might as well look at |x- 3|< 1. That is,-1< x- 3< 1 so 5< x+ 3< 7.