Question:
Prove that $\lim _{x \to \infty} x\sin x$ doesn't exist (using delta epsilon).
What I did:
I've been struggling with this one for a long time. Really tried digging up the net for explanations, but couldn't find a really good source to explain how this proof should be done(some said proof by contradiction like I did, and some say without contradiction). I got too confused by now, I also read this Proving limit doesn't exist using the $\epsilon$-$\delta$ definition
which gave some guidance but I thought I'd use the floor function to make sure M is natural, otherwise the idea of using these values wouldn't have worked.
Anyway I'd really love someone to verify/correct this solution before I hand it in.
Thanks a lot.
- We assume there exists a limit, therefore:
$\forall \varepsilon>0 \exists M>0: x>M \Rightarrow |f(x)-L|<\varepsilon$
and specifically, if we pick some $x_1,x_2>M$: $|f(x_1)-L|<\frac \varepsilon2,
|f(x_2)-L|<\frac \varepsilon2$ - Also, this applies for all $\varepsilon$, and specifically for $\varepsilon=0.5$
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(1) Therefore using the triangle inequality: $|f(x_1)-f(x_2)| = |f(x_1)-L+L-f(x_2)| \le |f(x_1)-L | + |f(x_2)-L| \le \frac \varepsilon2 + \frac \varepsilon2 = \varepsilon$
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This applies for all $x>M$ so specifically for: $x_1=\frac {\pi \lfloor M+1 \rfloor}{2} > M , x_2=\pi \lfloor M+1 \rfloor$
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We put these values in (1): $|f(x_1)-f(x_2)|=|\frac {\pi \lfloor M+1 \rfloor}{2}\sin\frac {\pi \lfloor M+1 \rfloor}{2}-\pi \lfloor M+1 \rfloor \sin (\pi \lfloor M+1 \rfloor)|= \begin{cases} 1 & \lfloor M+1 \rfloor odd \\ 0 &\lfloor M+1 \rfloor even\end {cases} $
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So we have found an $\varepsilon $ that doesn't always satisfy the requirements of the limit definition so this is a contradiction. Therefore the limit doesn't exist.
Best Answer
Solution by means of sequences:
$$x_n:=\pi n\implies x_n\sin x_n=0\xrightarrow[n\to\infty]{}0$$
$$y_n:=\frac{(4n-3)\pi}2\implies y_n\sin y_n=\frac{(4n-3)\pi}2\xrightarrow[n\to\infty]{}\infty$$