Let $\gamma_n : [a,b] \to \Bbb R^k$ a sequence of curves that converges uniformly to a curve $\gamma : [a,b] \to \Bbb R^k$. Do the lengths $l(\gamma_n)$ converge to the length $\gamma$ ? The answer is no with the counterexample $\gamma_n:[0,2\pi] \to \Bbb R^2$, $\gamma_n(x)=(x,\frac1n sin(n x))$ so that it converges to the null function on $[0,2\pi]$ but $l(\gamma_n)> 2\pi=l(\gamma)$. This seems counterintuitive to me and I would have answered yes at 100%. When I tried to prove it I was stuck at trying to show that $$\int_a^b \lVert \dot{\gamma_n(t)}- \dot{\gamma(t)} \rVert dt < \epsilon$$
which has nothing to do with $\lVert {\gamma_n(t)}- {\gamma(t)} \rVert dt < \epsilon$ for $n$ large enough. Can someone explain intuitively why does this length does not go to $0$ ? Is it because it has too many oscillations when $n$ goes to $\infty$ ? If the oscillations are barely getting away from $y=0$, the length should not change a lot...
Imagine we changed our curves to be $\gamma_n(x)=(x,\frac{1}{n}\sin(x))$. This is what they look like for $x\in[0,2\pi]$:
(this shows $\gamma_n(x)$ for $n=1,2,3,4$, and $n=10$ in purple)
It is clear that not only the curves approach the x axis uniformly but that their lengths approach $2\pi$ because the curves become more and more flattened out.
Now consider the original curves in your question, $\gamma_n(x)=(x,\frac{1}{n}\sin(nx))$:
They still approach the x axis uniformly, because for every two lines $y=\pm\epsilon$, we can find some $N$ such that for every $n>N$ the curves are between these two lines, as should be clear from the image.
However, the shape of the curve doesn't approach the shape of a straight line; they keep oscillating. This is because of the $\sin(nx)$ which makes them oscillate faster the higher $n$ is. Therefore, the lengths are always above $2\pi$ (and even above some $c>2\pi$), so they cannot converge to $2\pi$.