A function $f$ is continuous on a point ${x}_{0}$ if

For all $\epsilon >0$, there exists a $\phantom{\rule{thickmathspace}{0ex}}\delta >0$,
such that $\mid x-{x}_{0}\mid <\delta $ implies $\mid f(x)-f({x}_{0})\mid <\epsilon $.

Given a concret function (who is continuous), a ${x}_{0}$
and a $\epsilon >0$, you have to find a $\delta >0$
which verifies the above condition. And you will be noted according to
this $\phantom{\rule{thickmathspace}{0ex}}\delta $: more it is close to the best possible value, better
will be your note.
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