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Let $f:ℝ \to ℝ$ be a $C^∞$ curve. Determine the following limit;

$$\lim_{x_1 \to x_2} \dfrac{ \int_{x_1}^{x_2} \sqrt{1+f'(x)^2} dx}{\sqrt{(x_2-x_1)^2+(f(x_2)-f(x_1))^2}}$$

My attempt:

I recognized that the numerator represents the arc length of the curve from $P(x_1,f(x_1)$ to $Q(x_2,f(x_2))$ whereas the denominator is the length of the chord $PQ$. Intuitively, it seems as if when $Q$ tends to $P$, the numerator and denominator will become nearly equal so I think the limit should tend to $1$. I quickly verified this in case of a circle and it seems to work. However, I do not know how to prove this for any general function; applying L'hopital's rule once (after letting $x_2=x_1+h$) just complicates the limit further although it gets rid of the integral. And, I was looking forward to using this observation about arc length and chord length in my answer but I couldn't. Is there perhaps a better way of doing this using the squeeze theorem? I would appreciate any method which does not use L'hopital's rule and is not complicated computationally. Is my conjecture even correct?

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    $\begingroup$ Hint : write the denominator under the form $|x_2-x_1|\sqrt{1+\left(\frac{f(x_2)-f(x_1)}{x_2-x_1}\right)^2}$. $\endgroup$
    – Jean Marie
    Commented May 17 at 11:09
  • $\begingroup$ OH THAT'S THE DERIVATIVE $\endgroup$ Commented May 17 at 13:00
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    $\begingroup$ Just a picky comment. You want $\lim\limits_{x_1\to x_2^-}$. $\endgroup$ Commented May 17 at 19:30

1 Answer 1

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Writing the denominator under the form :

$$(x_2-x_1)\sqrt{1+\left(\frac{f(x_2)-f(x_1)}{x_2-x_1}\right)^2}$$

allows to write the quotient under the form $\frac{N}{D}$ where

$$N=\frac{1}{(x_2-x_1)} \int_{x_1}^{x_2} \sqrt{1+f'(x)^2} dx$$

and $$D=\sqrt{1+\left(\frac{f(x_2)-f(x_1)}{x_2-x_1}\right)^2}$$

Transformation of $N$ : Mean value theorem for integrals gives the existence of a certain $m$ with $x_1 \leq m \leq x_2$ such that : $$N=\sqrt{1+f'(m)^2}$$

Transformation of $D$ : Mean Value Theorem gives the existence of a certain $n$ with $x_1 \leq n \leq x_2$ such that :

$$\frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(n)$$

Can you conclude ?

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    $\begingroup$ You see we haven't use L'Hospital rule, which, rather often, can be a discovery method, but isn't advisable for rigorous proofs. $\endgroup$
    – Jean Marie
    Commented May 17 at 21:35

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