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Given a sequence of polygonal arcs $(f_n)_{n\in\mathbb{N}}$ that has limit $f$, is $f$ continuous?

I believe it should be true as I cannot think of a counter example but not sure how to prove it.

Thank you for your help.

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    @Chandrasekhar: but your sequence converges to $0$ in $[0,1]$, and the constant functions are continuous. What am I missing?2011-08-28

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Let $f_1$ be the polygonal arc defined thus. Let $P_0=(1,0)$, $P_1=(2/3,1)$, $P_2=(1/3,0)$, $P_\infty=(0,0)$. Join $P_0$ to $P_1$ to $P_2$ to $P_\infty$ by straight line segments.

Let $f_2$ be the polygonal arc defined thus. Let $P_3=(2/9,1)$, $P_4=(1/9,0)$. Join $P_0$ to $P_1$ to $P_2$ to $P_3$ to $P_4$ to $P_\infty$ by straight line segments.

In general, for $k \ge 1$, let $P_{2k-1} =(2/3^k,1)$ and $P_{2k}=(1/3^k,0)$, and let $f_n$ be the polygonal arc obtained by joining $P_i$ to $P_{i+1}$, for $i=0,1,\dots, 2n-1$, and finally $P_{2n}$ to $P_\infty$, by straight line segments.

We have imitated the behaviour of $\sin^2(1/x)$. The limiting curve is not continuous at $(0,0)$. If an endpoint difficulty is not acceptable, we can modify the definitions by adding to each polygonal arc $f_n$ its reflection across the $y$-axis.