Let $(f_n)_{n \in \mathbb{N}}$ be a sequence of functions $f_n \colon \mathbb{R} \to \mathbb{R}$ that converges uniformly to a function $f \colon \mathbb{R} \to \mathbb{R}$. By the uniform limit theorem, we know that if each $f_n$ is continuous, then for every point $x_0 \in \mathbb{R}$ we have
$$
\lim_{n \to \infty} \lim_{x \to x_0} f_n(x) = \lim_{x \to x_0} \lim_{n \to \infty} f_n(x).
$$
My question is: Does this also hold true for "$x_0 = \infty$" if both of the two limits exist?
I don't think this is true, and I tried some examples. But I could only find examples where one of the two limits doesn't exist at all, for example $f_n := \frac{x^2+nx}{n}$. If there are no examples where both limits exist as distinct real numbers, is there an example where one of the limits is $\pm \infty$ and the other is a real number?
Best Answer
It is true. Let $\epsilon >0$. There exists $n_0$ such that $|f_n(x)-f(x)| <\epsilon $ for $n \geq n_0$ for all $x$. If $L_n =\lim_{x \to \infty} f_n(x)$ and $L =\lim_{x \to \infty} f(x)$ the we can let $x \to \infty$ in above inequality to get $|L_n-L| \leq \epsilon$ for all $ n \geq n_0$. It follows that $L_n \to L$ which proves that $$ \lim_{n \to \infty} \lim_{x \to x_0} f_n(x) = \lim_{x \to x_0} \lim_{n \to \infty} f_n(x). $$