Define $f(x) = mx+b, \forall x$ and for fixed constants $m$ and $b$. Prove that the function $f : \mathbb{R} \to \mathbb{R}$ is uniformly continuous. I still have trouble with these epsilon delta proofs. Would we pick $\frac{\epsilon}{m}$? I achieved this by taking the abs value of $f(x)-f(y)$ and then dividing by $m$.
 A: Fix an $\epsilon>0$ now we need to find a $\delta>0$ that works for such $\epsilon>0$ and that $\delta>0$ never changes. 
We do things background 
$$|f(x) - f(y)| = |m(x - y)| < \epsilon \iff|x - y| < \frac{\epsilon}{|m|}. $$ Now pick $\delta := \frac{\epsilon}{|m|}$ and verify this indeed  works.
Here is the work:
$$|x - y| < \delta \implies |f(x) - f(y)|=|m||(x - y)|<|m|\delta=|m|(\epsilon/|m|)=\epsilon,$$ and you are done !
A: Well, let's check:
Pick any $x \in \mathbb{R}$, and we get:
$$|f(x) - f(x \pm \varepsilon/m)| = |mx + b - mx \pm \varepsilon - b| = |\pm \varepsilon| = \varepsilon $$
And so as desired, if we are within $\delta = \varepsilon/m$ distance in the inputs, then we will be within $\varepsilon$ distance in the outputs.  Of course, the $\delta$ isn't pulled out of a hat.  It looks like you had the right idea in finding it, and you'll find more detail in that regard in Karim's post.
You're good to go.
A: A more "principled" way to prove this might be to observe that:


*

*$f(x) = x$ is uniformly continuous;

*$f(x) = c$ is uniformly continuous for any constant $c$;

*If $f(x)$ is uniformly continuous and $c$ is a constant, then $c f(x)$ is uniformly continuous;

*If $f(x)$ and $g(x)$ are uniformly continuous, then $f(x) + g(x)$ is uniformly continuous.

If anyone is interested, we can also prove this using nonstandard analysis:
From Wikipedia,
we need to show that for hyperreal numbers $x,y$, if $x \approx y$, then $f(x) \approx f(y)$.
So let $y = x + \epsilon$ with $\epsilon$ infinitesimal, and we have
\begin{align*}
f(y) - f(x) &= f(x + \epsilon) - f(x) \\
&= \big[m(x + \epsilon)\big] + b - \big[mx + b\big] \\
&= m\epsilon \\
&\approx 0,
\end{align*}
since $\epsilon$ is infinitesimal and $m$ is finite.
A: You want to fix $\epsilon $ so that $$|f(c)-f(d)|=|m(c-d)| < \epsilon $$ 
When $|c-d| < \delta$ for a fixed $\delta$. Then you can use $|c-d|<....$
