"Suppose that $f:[0,\infty) \to \mathbb R$ is continuous and that there is an $L$ in $\mathbb R$ such that $f(x)\to L$ as $x \to \infty$. Prove that $f$ is uniformly continuous on $[0, \infty)$."
This is the proof my professor gave me to answer this question, but I don't understand what's going on after "if $|x-y| < \delta$" and the use of $M+1$, $M+2$. I believe $\delta$ has some relationship to the x values (as epsilon has some relationship to the y values); but is it true that for this reason we are moving up from M (which is used for infinity) so we add 1 when we are dealing with $x$ and add 2 when we are dealing with $y$?
I really appreciate any help and explanations. Thank you in advance!
It might help here to construct an incorrect proof, and then see why the above proof works :
For $\epsilon > 0$, there is $M \in \mathbb{R}$ such that $$ x > M \Rightarrow |f(x) - L| < \epsilon/2 $$ And so for $x,y > M$, one has $$ |f(x) - f(y)| < \epsilon $$ Also, $f$ is uniformly continuous on $[0,M]$, so there is $\delta > 0$ such that $$ x,y \in [0,M] \text{ and } |x-y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon $$ From this, you would think that $$ |x-y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon $$ should hold. Except, you need to account for the case where $x < M$ and $y>M$.
This leads to all the messing around with $M+1$ and $\delta \leq 1$, etc.