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I was asked by one of my student to prove the following problem:

Prove that the graph of the continous function $ y=f(x) = \left\{ \begin{array}{ll} \sqrt{x}\cos\left(\pi/x\right) & \quad 0 shown as

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is a nonrectible curve $C$.

I think, I should prepare a solid judgment based on discontinuity of $f'(x)$. What do you think? Please make me sure. Thanks for your time.

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    Discontinuity (or even non-existence) of $f'(x)$ alone won't cut it because $f(x) = |x-\frac12|$ certainly is rectifiable.2012-12-21

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I think it would be more convincing to estimate the length of each zig or zag its vertical height, and then show that the sum of these (under)estimates does not converge.

More precisely for $n\ge 2$, the length of the part of the curve between $x=\frac1n$ and $x=\frac{1}{n-1}$ is at least $\sqrt{\frac1n}$, and $\sum_n n^{-1/2}$ grows faster than the harmonic series.

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    Yes I see what I need to say her tomorrow . Thanks.2012-12-21
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Find a lower bound for the length which is easy to calculate. Solve $f(x)=0$ and $f(x)=+-\sqrt{x}$. Consider the poligonal made joining those points, show it's a lower bound. Calculate the length of this poligonal (it'll be an infinite sum instead of the original integral), in case it's infinite as the length of the curve is bigger it'll be infinite too.