# relation delta-epsilon within not exist limit

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 ?