Why does $f(x_i)-f(x_{i-1})$ not go to $0$ when finding arc length/surface area? I didn't really know what to title the question, but in class, we found the formula for surface area was: $A=\sum\limits_{i=1}^n\pi[f(x_i)+f(x_{i-1})]\sqrt{(x_i-x_{i-1})^2+(f(x_i)-f(x_{i-1}))^2}$ because surface area of a frustum is $A=\pi(R+r)l$ where $l$ is slant height. This makes sense to me but then the next step says that because $f(x_i)\approx f(x_i*)$ and $f(x_{i-1})\approx f(x_i*)$. Then they say $A=2\pi f(x_i*)\sqrt{1+f'(x_i*)^2}$. The square root was replace because of MVT which made sense to me, but if the approximation with $f(x_i*)$ is valid, then why wouldn't the square root just have the $f(x)$ part just go to $0$ and it'd just be $\Delta x$. If that substitution is invalid, then why does it work in the first part of the formula to give us $2f(x_i*)$.
Essentially what I'm asking is why are we able to say $f(x_i)\approx f(x_i*)$ and $f(x_{i-1})\approx f(x_i*)$ for the first part of the formula, but can't put those in for the part under the square root. I know this result would give us $\int2\pi f(x)dx$ which is wrong because it should be $ds$, but it feels wrong to apply the substitution in one place but not the other.
Thank you in advance for your help!
 A: Any derivation which merely uses $\approx$ is only meant to be heuristic, and I find them more confusing than a proper proof. This is not really informative because $\approx$ completely hides "how good" the approximation is supposed to be. Sure, you could definitely say
\begin{align}
A&=\sum_{i=1}^n\pi[f(x_i)+f(x_{i-1})]\sqrt{(x_i-x_{i-1})^2+[f(x_i)-f(x_{i-1})]^2}\\
&\approx\sum_{i=1}^n\pi[f(x_{i,*})+f(x_{i,*})]\sqrt{(x_i-x_{i-1})^2+[f(x_{i,*})-f(x_{i,*})]^2}\\
&=\sum_{i=1}^n2\pi f(x_{i,*})(x_i-x_{i-1})
\end{align}
But now the question becomes how good is this approximation actually? Is this approximation actually good enough so that when we take the limit as the size of the partition goes to zero, we have the limit of $A$ equals the limit of the sum on the last line? It turns out the answer is no. i.e we were way too crude and ignorant in the approximation $\approx$.
So, the way to argue this is not really just that $f(x_i)\approx f(x_{i-1})\approx f(x_{i,*})$. Rather, you have to estimate the difference between $A$, and a certain Riemann sum, and show that that difference is negligible as the size of the partition goes to $0$. Explicitly, using the mean-value theorem on each interval $[x_{i-1},x_i]$, we get a point $\xi_i$ (I don't want to write $x_{i,*}$ simply for easy of typing). Let me define
\begin{align}
R&:=\sum_{i=1}^n2\pi f(\xi_i)\sqrt{1+f'(\xi_i)^2}(x_i-x_{i-1})
\end{align}
This is a proper (tagged) Riemann sum. Now, let us show that $|A-R|$ can be made small provided that the size of the partition goes to $0$:
\begin{align}
|A-R|&=\left|\sum_{i=1}^n\pi[f(x_i)+f(x_{i-1})]\sqrt{1+f'(\xi_i)^2}(x_i-x_{i-1})-\sum_{i=1}^n2\pi f(\xi_i)\sqrt{1+f'(\xi_i)^2}(x_i-x_{i-1})
\right|\\
&=\left|\sum_{i=1}^n\pi [f(x_i)+f(x_{i-1})-2f(\xi_i)]\sqrt{1+f'(\xi_i)^2}(x_i-x_{i-1})\right|\\
&\leq \sum_{i=1}^n\pi \left[|f(x_i)-f(\xi_i)|+|f(x_{i-1})-f(\xi_i)|\right]\sqrt{1+f'(\xi_i)^2}(x_i-x_{i-1})
\end{align}
Suppose you're working on a compact interval $[a,b]$. Then using uniform continuity of $f$ on $[a,b]$, given $\epsilon>0$ there is a $\delta>0$ such that for all $p,q\in [a,b]$, if $|p-q|\leq \delta$ then $|f(p)-f(q)|\leq \epsilon$. So, by choosing partitions $P=\{x_0,\dots, x_n\}$ with mesh at most $\delta$, it follows that for each term, $|x_i-\xi_i|\leq \delta$ and $|x_{i-1}-\xi_i|\leq \delta$. So, in the above sum, the differences in the values of $f$ at the two points is bounded by $\epsilon$:
\begin{align}
|A-R|&\leq \sum_{i=1}^n\pi [\epsilon+\epsilon]\sqrt{1+f'(\xi_i)^2}(x_i-x_{i-1})\\
&\leq \sum_{i=1}^n2\pi \epsilon\sqrt{1+M^2}(x_i-x_{i-1})\\
&=2\pi\sqrt{1+M^2}(b-a)\epsilon,
\end{align}
where $M:=\sup\limits_{t\in [a,b]}|f'(t)|$ is an upper bound on the derivative of $f$ (to make sense of this derivation, one has to impose certain regularity assumptions on the function, so this is justified). THis last inequality is just a constant multiple of $\epsilon$.
Therefore, it follows that the limit of the area approximations and of these specific RIemann sums are the same. In symbols (actually the above $\epsilon$-$\delta$ formulation is the definition of these limit symbols),
\begin{align}
\lim_{\|P\|\to 0}A&=\lim_{\|P\|\to 0}R=\int_a^b2\pi f(x)\sqrt{1+f'(x)^2}\,dx
\end{align}

On the other hand, if you consider the following sum
\begin{align}
S&=\sum_{i=1}^n2\pi f(\xi_i)(x_i-x_{i-1})
\end{align}
then you can't establish the same kind of inequality as we did above, i.e we can't establish an inequality of the form $|A-S|\leq \text{constant}\cdot \epsilon$ (and the indirect way of proving this is that previously, I already provided a correct proof that $R$ is the correct approximation, so by uniqueness of limits, these $S$ sums aren't the correct approximation). Thus, this approach is doomed.

Very roughly speaking, the estimation which uses the MVT (i.e which yields the Riemann sums $R$) used a "first order approximation" (i.e involves derivatives $f'(\xi_i)$) followed by a "zeroth order approximation" (i.e using continuity... ok actually I used uniform continuity). However, if you use only the approximation $f(x_i)\approx f(x_{i-1})\approx f(\xi_i)$, this is only continuity of $f$ (zeroth-order). This is a much weaker estimation, and unfortunately it's just not good enough.
A: Suppose that $x_0=a$, $x_n=b$ are constant and $x_0<x_1<x_2<...<x_n$. Consider limit
$$A=\lim_{\max(x_{i}-x_{i-1})\to 0, n\to \infty}\sum\limits_{i=1}^n\pi[f(x_i)+f(x_{i-1})]\sqrt{(x_i-x_{i-1})^2+(f(x_i)-f(x_{i-1}))^2}$$
To express this limit as
$$\lim_{\max(x_{i}-x_{i-1})\to 0, n\to \infty}\sum\limits_{i=1}^n g(x_i)\cdot (x_i-x_{i-1})=\int_a^b g(x)\,dx$$
we need to factor out $x_i-x_{i-1}$, then
$$g(x_i)=\lim_{x_i-x_{i-1}\to 0}\frac{\pi[f(x_i)+f(x_{i-1})]\sqrt{(x_i-x_{i-1})^2+(f(x_i)-f(x_{i-1}))^2}}{x_i-x_{i-1}}$$
$$g(x_i)=\lim_{x_{i-1}\to x_i}\frac{\pi[f(x_i)+f(x_{i-1})]\sqrt{(x_i-x_{i-1})^2+(f(x_i)-f(x_{i-1}))^2}}{x_i-x_{i-1}}$$
$$g(x_i)=2\pi f(x_i)\lim_{x_{i-1}\to x_i}\sqrt{1+\left(\frac{f(x_i)-f(x_{i-1})}{x_i-x_{i-1}}\right)^2}$$
$$g(x_i)=2\pi f(x_i)\sqrt{1+\left(\lim_{x_{i-1}\to x_i}\frac{f(x_i)-f(x_{i-1})}{x_i-x_{i-1}}\right)^2}$$
$$g(x_i)=2\pi f(x_i)\sqrt{1+f'(x_i)^2}$$
