Prove $mx+b$ is continuous at any point in $\mathbb{R}$

Question

I need to prove, $mx+b$ is continuous at any point in $\mathbb{R}$

Now, as I have thought of there's 2 possible cases:
1) $m = 0$
2) $m \neq 0$

So for case #2, $m < 0 \vee m > 0$ ,
and we can assume without loss of generality $m>0$

$|x-a| < \delta \longrightarrow |f(x) - f(a)| < \epsilon $
$|x-a| < \delta \longrightarrow | (mx+b) - (ma+b) | < \epsilon$
$|x-a| < \delta \longrightarrow |mx-ma| < \epsilon $
$|x-a| < \delta \longrightarrow |x-a| < \epsilon/m $

And then, let $\delta =\epsilon/m $
$|x-a| < \delta \longrightarrow |x-a| <\epsilon/m $
$\longrightarrow |mx -ma| < \epsilon$
$\longrightarrow |mx +b -ma -b|\epsilon$
$\longrightarrow |f(x) - f(a)|\epsilon$
Thus $mx+b $ is continuous when $m > 0 \vee m<0$

But, now how do I prove it when $m = 0$?

I end up with $\epsilon/m$ and $m = 0$ and that is no correct.

So, how should I prove $mx+b$ is continuous when $m = 0$ OR am I approaching the problem wrong?

Thanks in advance for all the help! :D

Answer 1

Pick a point $a \in \mathbb{R}$. Let $\epsilon >0$. Choose $\delta=\displaystyle\frac{\epsilon}{|m|}$. Now let $|x-a|<\delta$. We have,

\begin{align*} |(mx+b)-(ma+b)|&=|m||x-a|<|m|\delta=\epsilon\\ \end{align*}

Therefore, $f(x)=mx+b$ is continuous at $a$. $a$ was arbitrary so continuity holds everywhere.