Using $\frac{\epsilon}{3}$ to show a limit implies uniform continuity?

Question

If $f:[0,\infty)\to\mathbb{R}$ is continuous and $\lim_{x\to\infty}f(x)=L<\infty$, then $f$ is uniformly continuous on $[0,\infty)$.

Using the formal definition of a limit with $\frac{\epsilon}{3}$ to obtain a positive number $M$ such that $f(x)$ is within $\frac{\epsilon}{3}$ to $L$ for all $x\geq M$.

I need to know the case when $x\in[0,M]$ and $y\in(M,\infty)$. Show the definition of continuity holds in this case using $\frac{\epsilon}{3}$.

Answer 1

For any $\epsilon$, let $N$ be large enough such that for $x\ge N$, you have $|f(x)-f(N)|<\frac{\epsilon}{3}$. Since $f(x)$ is continuous on $[0,N]$, we may find $\delta$ such that for $|x-y|<\delta$, $|f(x)-f(y)|<\frac{\epsilon}{3}$ as well. Now for any $x,y,d(x,y)<\delta$, if $x\le N\le y$ we should have

$$|f(x)-f(y)|\le |f(x)-f(N)+f(N)-f(y)|\le |f(x)-f(N)|+|f(N)-f(y)|< \frac{\epsilon}{3}+\frac{\epsilon}{3}=\frac{2\epsilon}{3}<\epsilon$$ which showed $f$ to be uniformly continuous on $\mathbb{R}$. As Julien commented $1/3$ can be changed to $1/2$ or $1/2,1/4$, etc.