Let $f_n\to f$, uniformly on $[0,\infty)$ and let's assume that $\lim_{x\to\infty} f_n(x) = \alpha_n$ and $\alpha_n\to \alpha$. Show that: $$\lim_{x\to\infty} f(x) = \alpha$$
Now, the proof goes like this.
- Let $\varepsilon > 0$.
- We take $N_0$ such that $\sup |f_n(x) - f(x)| \le \varepsilon /3$ for all $n\ge N_0$.
- We take $N_1$ such that $|\alpha_n - \alpha| \le \varepsilon /3$ for all $n\ge N_1$.
- We take $N = \max(N_0,N_1)$
- Then, there's an $\color{red}{M>0}$ such that $\color{red}{|f_N(x)-\alpha_N(x)| \lt \varepsilon / 3}$, for all $\color{red}{x\ge M}$.
Therefore, for all $x\ge M$:
$$|f(x)-\alpha| \le |f(x)-f_n(x)| + |f_n(x)-\alpha_n| + |\alpha_n -\alpha| \le \varepsilon$$
Now, I don't fully understand the red part. Why there's such $M>0$?