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$a_n$ is cauchy sequence on (0,1) I have two question ,

1.) If $f:[0,1]\to\mathbb R$ and $b_n=\int_0^{a_n}f(x)$ then $b_n$ is cauchy sequence.

2.) If $g$ is differentiable on (0,1) and $c_n=g'(a_n)$ then $c_n$ is cauchy sequence.

I know $b_n$ and $c_n$ are cauchy sequence if and only if $b_n$ and $c_n$ are convergent sequence,

but i don't know to show that please give me a hint or solve this.

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  1. Assuming that $f$ is Riemann-integrable, yes. Because then $|f|$ is bounded by some $M$ and then$$|b_m-b_n|=\left|\int_0^{a_m}f(x)\,\mathrm dx-\int_0^{a_n}f(x)\,\mathrm dx\right|\leqslant M.|a_m-a_n|.$$
  2. Not in general. Suppose that $g(x)=\sin\left(\frac1x\right)$ and that $a_n=\dfrac1{\frac\pi2+n\pi}$.
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  • $\begingroup$ I try $g'(x)=-\frac1{x^2}cos(\frac 1x)$ and $c_n=g'(a_n)$ then show $lim_{n\to \infty} c_n$ is not exist ? $\endgroup$ – Aong Wa Mar 30 '18 at 12:44
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Hint:

$$|b_n-b_m|=\left |\int_{a_m}^{a_n}f(x)dx \right|$$ what do you know about $|a_n-a_m|$?

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  • $\begingroup$ i know $|a_n-a_m|<\epsilon$ then $a_n$ is cauchy sequence. $\endgroup$ – Aong Wa Mar 30 '18 at 12:46

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