What is $\lim_{n\to \infty }\Gamma _n$ where $\Gamma _n$ is a set. Consider the family of curve $\Gamma _n$ as in the following graphic
 
So $\Gamma _0$ is in black, $\Gamma _1$ in red, $\Gamma _2$ in blue, and we continue like this. We have that $$\Gamma _n=\{(t,\varphi _n(t))\mid t\in I\}$$
for a certain function $\varphi _n:I\to \mathbb R$, where $I$ is an interval. It's clear that $\varphi _n(t)\to 0$ uniformly, but clearly $$\lim_{n\to \infty }\Gamma _n\neq \left\{\lim_{n\to \infty }(t,\varphi _n(t))\mid t\in I\right\}=\{(t,0)\mid t\in I\},$$
since $\ell\left(\lim_{n\to \infty }\Gamma _n\right)=\pi$ whereas $\ell\left(\{(t,0)\mid t\in I\}\right)=2$ (where $\ell(A)$ denote the length of the curve $A$). So, what could be $$\lim_{n\to \infty }\Gamma _n \ \ ?$$
Is it possible to describe it ?
 A: We have $$\lim_{n\to \infty }\Gamma _n=\{(t,0)\mid t\in D\},$$
where $$D=\bigcup_{n\in\mathbb N}D_n,$$
and $$D_n=\left\{\frac{k}{2^n}\mid k\in \{-2^n,-2^n+1,...,2^n-1,2^n\}\right\}.$$
Remark that $$\varphi _n(t)=0\iff t\in D_n.$$
Let $x\in \limsup_{n\to \infty }\Gamma _n$. In particular, for all $n$ there is $m_n\geq n$ s.t. $x=(t,\varphi _{m_n}(t))$. Taking $n\to \infty $ yields $x=(t,0)$. Therefore, there is $n\in \mathbb N$ s.t. $\varphi _n(t)=0$, i.e. $t\in D$. In particular, since $D_m\subset D_n$ for all $m\geq n$, we get that $\varphi _m(t)=0$ for all $m\geq n$. Therefore $x=(x,\varphi _m(x))$ for all $m\geq n$ and thus $x\in \liminf_{n\to \infty }\Gamma _n$. This prove that $$\liminf_{n\to \infty }\Gamma _n=\limsup_{n\to \infty }\Gamma _n\subset \{(t,0)\mid t\in D\}.$$
For the converse inclusion, if $t\in D$, then there is $n\in\mathbb N$ s.t. $t\in D_n$, and thus $\varphi _{m}(t)=0$ for all $m\geq n$. Therefore $(t,0)\in \liminf_{n\to \infty }\Gamma _n,$ and thus, the claim follow.

And indeed, as you correctly remark, the limit curve $\lim_{n\to \infty }\Gamma _n$ is different than $\{\lim_{n\to \infty }(t,\varphi _n(t))\mid t\in I\}$. 
