Proving discontinuity by epsilon-delta

We have $f(1)=5$. So to show that $f$ is not continuous at $x=1$, it is enough to show that it is not true that $\lim_{x\to 1} f(x)= 5$.

Suppose to the contrary that the limit exists and is equal to $5$. Then for any $\epsilon\gt 0$, there is a $\delta\gt 0$ such that if $|x-1|\lt\delta$, then $|f(x)-5|\lt\epsilon$.

Pick $\epsilon=\frac{1}{2}$. We show there is no $\delta$ with the required property.

If $x\lt 1$, then $f(x)\lt 1$. In particular, if $x\lt 1$, then $|f(x)-5|\gt 4$. It follows that there is no $\delta$ such that $|x-1|\lt \delta$ guarantees that $|f(x)-5|\lt \epsilon$.

Remark: The idea is basically geometric. We are showing that there are points arbitrarily near $x=1$ at which the function value is not close to $5$. Once one has the geometry under control, writing out the details in $\epsilon$-$\delta$ language is a translation job.