A continuous functions $f: [a,\infty) \to \mathbb R$ is uniformly continuous if $\;\lim_{x \to \infty} f(x) = L,\;L,a \in \mathbb R$
I want to show my proof here for verification. If there is any mistake in my proof, I would appreciate it if you would point that out.
Given an arbitrary $\epsilon >0$, we have to show that there exists a corresponding $\delta >0$ such that $|x-y|< \delta \implies |f(x) - f(y)|<\epsilon\quad \forall x,y \in [a, \infty)$.
By Heine-Cantor theorem, for any compact interval like $[a,t]$, there exists a $\;\delta_t>0\;$ such that $|x-y| < \delta_t \implies |f(x)-f(y)|< \epsilon\quad x,y \in [a,t]$.Since $\lim_{x \to \infty} f(x) = L$, by definition, there exists a $M>0$ such that $x \geq M \implies |f(x)-L|< \epsilon/2\;$. we also choose $\alpha>0$ in a way that $|x-M|<\alpha\implies |f(x) - f(M)| < \epsilon/2$; by continuity, such a choice for $\alpha$ is possible. Put $\delta$ = min{$\delta_M, \alpha$}. we claim that this is the desired $\delta$.
case1: $x,y \geq M$. In this case, $|f(x)-f(y)| \leq |f(x) - L| + |f(y)-L| < \epsilon/2 + \epsilon/2=\epsilon$
case2: $x,y \leq M$. In this case, $|x-y|< \delta< \delta_M \implies |f(x)-f(y)|<\epsilon$
case3: $x \leq M$ and $ y \geq M$. $|x-y| < \delta < \alpha \implies |x-M|< \alpha$ and $|y-M|< \alpha$. $|f(x)-f(y)| \leq |f(x) - f(M)| + |f(M)- f(y)| \leq \epsilon/2 + \epsilon/2 = \epsilon$
Since $\epsilon$ was arbitrary, the proof is complete.