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I have this: if $f:[1, \infty)\to\mathbb{R}$ and the limit $\int^\infty _1 f(x)dx$ exists then $\lim\limits_{x\to \infty}f(x)=0$

How can I show this to be true, is it similar if it were $[0,\infty)$? Or am I incorrect in thinking this statement to be true?

Thanks to anyone, who can help!

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  • $\begingroup$ Do you need the limit to exist? $\endgroup$ Commented Feb 1, 2023 at 15:49
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    $\begingroup$ It is well known that, unlike sequences, the existence of $\int_1^{+\infty}f(x)dx$ doesn't imply that $\lim\limits_{x\rightarrow +\infty}f(x)=0$. The lower bound in the integral is not important because we are only interested in the behaviour of $f$ at infinity. $\endgroup$
    – Tuvasbien
    Commented Feb 1, 2023 at 15:50
  • $\begingroup$ Related: math.stackexchange.com/questions/102678/… $\endgroup$ Commented Feb 1, 2023 at 15:52
  • $\begingroup$ Also Related: math.stackexchange.com/questions/109826/… $\endgroup$ Commented Feb 1, 2023 at 15:54
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    $\begingroup$ Your statement is syntactically unclear (there are 2 "then"). So is your title (there are 2 "the"). $\endgroup$ Commented Feb 1, 2023 at 16:00

1 Answer 1

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A counter example: $$ \int_1^\infty \sin(x^2)dx $$ exists. However $$ \lim_{x\to\infty}\sin(x^2) $$ doesn’t exist.

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