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Is it true that

$$\displaystyle\lim_{x\to 0}x\int_{0}^\infty k^5 dk=0$$

Seems like this has an indeterminate form, but I am being told it equals zero.

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    $\int^\infty_0 k^5~\mathrm{d}k$ is usually not defined. – Hyperbolic PDE friend Jun 09 '22 at 19:26
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    It would be useful to know who told you that and what their arguments are. $\int_0^\infty k^5dk$ is infinite, so you cannot even multiply it by $x$, let alone take the limit! So for me it is almost "case closed" but I would like to give the person who told you it was zero the benefit of the doubt. (Or, more likely, correct their misunderstanding.) –  Jun 09 '22 at 19:26
  • @StinkingBishop It was a tutor. Their argument was along the lines that the integral can be represented as a series, and since each individual term in the series multiplied by x goes to zero, the whole integral does as well. –  Jun 09 '22 at 19:30
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    It is true that in some cases you can swap limit and integral. If the argument is $\lim_{x\to 0}x\int_0^\infty k^5dk=\lim_{x\to 0}\int_0^\infty xk^5dk\overset{???}{=}\int_0^\infty (\lim_{x\to 0}xk^5)dk=\int_0^\infty 0\cdot dk=0$, then the problem is to justify the step $\overset{???}{=}$. In this case it is not justified. Have a look at e.g. https://math.stackexchange.com/a/253697/700480 for some criteria. –  Jun 09 '22 at 19:33
  • (And, in fact here you have two limits: $\lim_{x\to 0}\lim_{a\to\infty}\int_0^a xk^5dk$ and the "outer" limit needs to "skip over" both the inner limit and the (proper) integral. $\int_0^a$.) –  Jun 09 '22 at 19:37
  • Thank you for the explanation, this clears up some confusion on my end! –  Jun 09 '22 at 19:42
  • A simpler example where you have a sequence of series. They are all convergent, sum of each is $1$ but each individual term converges ultimately to $0$. Look at this: $1+0+0+0+\ldots, 0+1+0+0+\ldots, 0+0+1+0+\ldots, \ldots$, so the hand-wavy argument that if individual terms tend to $0$ the whole sum needs to tend to zero just doesn't work. Again, there are cases when it does (e.g. when the convergence is uniform) but the onus would be on your tutor to recall and use the relevant theorems in the proof (if any). –  Jun 09 '22 at 19:45

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