Length of a Continuous Curve Let f be a bounded real valued function on [a,b], then define the length of the curve L(f) as follows:
\begin{equation}\tag{1}
L(f)=sup\sum\limits_{i=1}^{n}\sqrt{ (x_i-x_{i-1})^2+(f(x_i)-f(x_{i-1}))^2}, 
\end{equation}
where sup is taken over all partitions of [a,b]. 
The question is to show, L(f) is equal to
\begin{equation}\tag{2}
\lim\limits_{max|x_i-x_{i-1}|\to 0}\sum\limits_{i=1}^{n}\sqrt {(x_i-x_{i-1})^2+(f(x_i)-f(x_{i-1}))^2}
\end{equation}
for continuous functions.
I know when $max|x_i-x_{i-1}|\to 0$, we get finer partitions so that $(1)\leq(2)$ because if we pick any partition and then let $max|x_i-x_{i-1}|\to0$ then we get a greater sum due to triangle inequality.(I guess what I said is true and does not require continuity.)
I could not show  $(2)\leq(1)$ and could not use the continuity. In a finite sum, we can switch $\sum$ and limit but as $max|x_i-x_{i-1}|\to0$ we get an infinite sum so switching may fail. I got stuck.
P.S. I have read (Determining the Length of a Curve Using Partitions) but I did not understand "there is a sequence $(Q_n)_{n\geq0}$ of partitions with (3) holds" which was stated in the answer. If you refer to that question, please tell me why (3) holds. I thought that (3) is the same as what we want to show in that question. 
Probably, I missed an easy detail. Thanks in advance for any help. 

The basic fact you are (or should be) using is that if $S$ is any set of real numbers, and $X=x_1, \dotsc, x_n, \dotsc$ is any convergent subsequence, then $L=\lim X \leq \sup S.$ This is obvious, because if $L > \sup S,$ then there is an $i$ such that $x_i \geq L - (L-\sup S),$ contradicting the supremum property.
