I am to show the continuity of this function with the help of $\epsilon$-$\delta$ argument.
The function is: $g: \Bbb{R} \rightarrow \Bbb{R}$, $x \mapsto x^2$.
Given the $\epsilon$-$\delta$ definition of limit, I tried as follows:
We must have: $|f(x)-f(x_0)|<\epsilon$, so then $|x^2-x_0^2|=|(x-x_0)(x+x_0)|=|(x-x_0)||(x+x_0)|$, so $|x-x_0|<\frac{\epsilon}{|x+x_0|}$. So now I will choose $\delta=\frac{\epsilon}{|x+x_0|}$.
Is it enough? I am writing this sentences like a machine, but I am not understanding intuitively.