In this post, @Post No Bulls claims that $$ f(x)=\sum_{n=1}^\infty \max(0, 1-2^n|x-n|) $$ is square integrable on $\mathbb R$ but $\limsup_{x\to\infty} f(x)=1$, and $\lim_{x\to\infty} f(x)$ does not exist.
How to prove that $f$ is square integrable on $\mathbb{R}?$
The intervals $(n-\frac 1 {2^{n}},n+\frac 1 {2^{n}})$ are disjoint. A particular point $x$ can belong at most one of these. When $x$ is in $(n-\frac 1 {2^{n}},n+\frac 1 {2^{n}})$ the value of $f(x)$ lies between $0$ and $1$. Hence $\int f(x)^{2} dx \leq \sum \frac 2 {2^{n}} <\infty$.
$0 \leq f \leq 1$, $f(n)=1$ and $f(n+\frac 1 {2^{n}})=0$ for all $n$. Hence $\lim\sup_{x \to \infty} f(x)=1$ and $\lim_{x \to \infty} f(x)$ does not exist.