Let $\{f_{n}:\mathbb{R}\to\mathbb{R}\}$ be a sequence of functions with uniform limit $f:\mathbb{R}\to\mathbb{R}$, and let $\{c_{n}\}$ be a sequence defined by $$c_{n}:=\lim_{x\to\infty}f_{n}(x),$$ where the limit $\displaystyle\lim_{x\to\infty}f_{n}(x)$ exists in $\mathbb{R}$ for each $n\in\mathbb{N}$. Prove that $\{c_{n}\}$ is a convergent sequence.
The following are my two attempts:
(1) Since $\{f_{n}\}$ converges uniformly to $f$ on $\mathbb{R}$, for any $\varepsilon>0$, there exists $K\in\mathbb{N}$ such that $$|f_{m}(t)-f_{n}(t)|<\frac{\varepsilon}{3}$$ whenever $m,n\ge K$ and $t\in\mathbb{R}$.
Now, for each $n\in\mathbb{N}$, there exists $t_{n}\in\mathbb{R}$ large enough so that $$|f_{n}(x)-c_{n}|<\frac{\varepsilon}{3}$$ whenever $x\ge t_{n}$.
So, for any $m,n\ge K$, choose $a=\max\{t_{m},t_{n}\}$, then $$|c_{m}-c_{n}|\le|c_{m}-f_{m}(a)|+|f_{m}(a)-f_{n}(a)|+|f_{n}(a)-c_{n}|<\varepsilon,$$ i.e., $\{c_{n}\}$ is a Cauchy sequence in $\mathbb{R}$. Hence, $\{c_{n}\}$ is a convergent sequence. $\square$
(2) Since $\{f_{n}\}$ converges uniformly to $f$ on $\mathbb{R}$, for any $\varepsilon>0$, there exists $N\in\mathbb{N}$ such that $$|f_{m}(x)-f_{n}(x)|<\varepsilon$$ whenever $m,n\ge N$ and $x\in\mathbb{R}$.
So, if $m,n\ge N$, we have $$|c_{m}-c_{n}|=\left|\lim_{x\to\infty}f_{m}(x)-\lim_{x\to\infty}f_{n}(x)\right|=\lim_{x\to\infty}\left|f_{m}(x)-f_{n}(x)\right|\le\varepsilon.$$
The first proof seems to be no problem. However, the second proof doesn't seem rigorous.
Can anyone criticize my proofs? Give some advice. Thank you!
I do not think that the statement is true at all,even if continuity is assumed.
Take the sequence $f_n=x+\frac{1}{n}$
It converges uniformly to $x$
But $c_n=+\infty,\forall n \in \Bbb{N}$ as $x \to +\infty$.
If each $f_n$ was bounded for every $n \in \Bbb{N}$ or more strongly,the sequence was uniformly bounded then your attempts are correct.