Let f be a function such that f(x)=x^2 for all numbers x. Show that f is continuous at the point (2,4).
Well you want it to work.
$\displaystyle \left| {x - 2} \right| < \delta = \min \left\{ {1,\varepsilon /5} \right\}\, \Rightarrow \,\left| {x + 2} \right| < 5$
$\displaystyle \left| {x^2 - 4} \right| = \left| {x + 2} \right|\left| {x - 2} \right| < 5\left( {\varepsilon /5} \right) = \varepsilon $
I think that you need to do some background work on this.