We safe $f$ is continuous at $x_0$ if for any $\epsilon >0$ there exists $\delta > 0$ such that whenever $|x-x_0|<\delta$ then $|f(x)-f(x_0)|<\epsilon$.
I'm reviewing for my final, and I have been coming up with functions and trying to show they are continuous. Here are the two I came up with: (1) $f(x)=x^2$ and (2) $f(x)=x^2-9$
To show (1) is continuous: $|f(x)-f(2)|=|x^2-4|=|x+2||x-2|$ Let $|x-2|<1$, then $1
If we let $\delta=\min\{1,\epsilon/5\}$ we get what we wanted.
To show (2) is continuous at $x=2$ is a bit trickier I believe since it doesn't factor into pieces as evenly. This is where I'm stuck. Is there a way to rearrange the terms to get it to factor more nicely?