I'm getting really curious about the limit

$$\lim_{x \to \infty}x\cdot o\left (\frac{1}{x^2} \right) $$

Well it's pretty obvious that $o\left (\frac{1}{x^2} \right) \rightarrow 0$ and $x \rightarrow \infty$ when $x \rightarrow \infty$. So I get $\infty \cdot 0$ which is undefined. But maybe I'm wrong? Maybe there appares some kind of mathematical 'magic' and the limit is $0$ or $\infty$?

Does anyone know anthing about that?

  • $\begingroup$ What 's $o(x)$? $\endgroup$ – Your IDE Dec 15 '17 at 11:47
  • $\begingroup$ $o \left( \frac{1}{x^2} \right)$ is a little- $o$ notation $\endgroup$ – Karagum Dec 15 '17 at 11:49
  • $\begingroup$ The function is certainly bounded above by $x/x^2=1/x$ and you can conclude. $\endgroup$ – Yves Daoust Dec 15 '17 at 12:32

tl;dr: $$ x\cdot o\left(\frac{1}{x^2}\right) = o\left(\frac{1}{x}\right) $$

To see why this is the case, if you don't see it, go back to the definition.

$f(x) = o\left(\frac{1}{x^2}\right)$ when $x\to\infty$

means that $\lim_{x\to\infty} x^2f(x) = 0$. So here, "hidden" in the $o(\cdot)$ notation we have some function $f$ such that $$ x\cdot o\left(\frac{1}{x^2}\right) = xf(x)\,. $$ write $g(x) = xf(x)$: we have $\lim_{x\to\infty}xg(x) = \lim_{x\to\infty}x^2 f(x) = 0$ by assumption, that is to say $g(x) = o\left(\frac{1}{x}\right)$ by definition.


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