Im having a bit of trouble proving a function is continuous for all reals except for one point. I can do the proof for one point but I'm having trouble extending the proof to all reals except one point.
If $\varphi(x)$ is a function from $\mathbb{R}$ to $\mathbb{R}$ such that $\varphi(x)=0$ for $x\neq5$ and $\varphi(5)=3$, prove that $\varphi$ is continuous at every $k\neq5$ using the epsilon delta definition.
On
Obviously $\phi$ is not continuous at $x=5$ since for $\epsilon = 1$, $\vert \phi(5) - \phi(5 + \delta) \vert > 1$ for all $\delta \neq 0$. By taking $\delta > 0$ is it not right-continuous at $x=5$, and taking $\delta < 0$ shows that it is not left-continuous either.
Now fix any $x \neq 5$ and suppose that $\vert 5 - x \vert = y$. Then, fix $\epsilon > 0$ and note that regardless of the choice of $\epsilon$, taking $\vert\delta\vert < y$ gives $\vert \phi(x) - \phi(x + \delta) \vert = 0$ since $x + \delta$ cannot be $5$. Hence, $\phi$ is continuous at all $x \neq 5$.
Suppose $x\not=5$. Then there exists an $r>0$ such that $(x-r,x+r)\subseteq\mathbb{R}\setminus\{5\}$. Now take any $\epsilon$ with $0<\epsilon<r$ and let $\delta := r$. Then for any $y\in(x-\delta,x+\delta)$ we have that $|\varphi(y) - \varphi(x)| = |0-0|=0<\epsilon$. Thus $\varphi$ is continuous at $x$.
Note that the epsilon delta condition holding for all $\epsilon>0$ is equivalent to it holding for all $c>\epsilon>0$ where you can pick $c$ to be any positive real number. This is a neat little trick which often comes in handy.