= Solution
The function is <continuous function>[continuous] at $x$ when, for every $\varepsilon>0$, there is $\delta>0$ such that $|y-x|<\delta$ implies $|f(y)-f(x)|<\varepsilon$.
If this holds and $x_n\to x$, eventually $|x_n-x|<\delta$, so $f(x_n)\to f(x)$. Conversely, if continuity fails, some $\varepsilon_0>0$ allows, for every $n$, a point $x_n$ with $|x_n-x|<1/n$ but $|f(x_n)-f(x)|\geq\varepsilon_0$. Then $x_n\to x$ while $f(x_n)\not\to f(x)$, a contradiction. This is the <sequential characterization of continuity in metric spaces>.
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