when we are proving limit that supposed to be exist , we are have a nice relation between $\epsilon$ and $\delta$. i mean regardless what epsilon we choose we will able to find a suitable delta , that later will help us to find parameter for how should we choose x , but if the limit is not exist, how both epsilon and delta related? here is one example and it's solution and i have something to ask about it,

Q: prove that $\lim _{x \to \infty} x\sin x$ doesn't exist

A: assume the limit exists and is $L$ fix $\epsilon = 1$ . there exist $M > 0$ such that for $x > M .| x \sin x - L | < \epsilon = 1$ , set $P = max \{L , M\}$ . let $ x_0 = (\frac {\pi} {2} +2P\pi)$ we have , $x_0 > M$ and $x_0 \sin x_0 = (\frac {\pi} {2} +2P\pi) → | x_0 \sin x_0 - L | > 1$ hence we have the contradiction

so, my first question, since $\epsilon$ fixed by 1 how it relation with $M$ that we can choose $x_0= (\frac {\pi} {2} +2P\pi)$ ? in order follow $x > M $. i know if the limit exist we can express epsilon using delta for example $\epsilon = \sqrt {\delta} or \epsilon = \frac {1} {\delta}$.something like that but how in this case ?

second, in " set $P = max \{L , M\}$ " part, i still understand to choose P=M because surely $$(\frac {\pi} {2} +2M\pi) > M$$ reflect to $x_0 > M $. but why there is L ?


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