# Prove that $\lim _{x \to \infty} x\sin x$ doesn't exist (using delta epsilon).

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 $\varepsilon=0.5$
• (1) Therefore using the triangle inequality: $|f(x_1)-f(x_2)| = |f(x_1)-L+L-f(x_2)| \le |f(x_1)-L | + |f(x_2)-L| \le \frac \varepsilon2 + \frac \varepsilon2 = \varepsilon$

• This applies for all $x>M$ so specifically for: $x_1=\frac {\pi \lfloor M+1 \rfloor}{2} > M , x_2=\pi \lfloor M+1 \rfloor$

• We put these values in (1): $|f(x_1)-f(x_2)|=|\frac {\pi \lfloor M+1 \rfloor}{2}\sin\frac {\pi \lfloor M+1 \rfloor}{2}-\pi \lfloor M+1 \rfloor \sin (\pi \lfloor M+1 \rfloor)|= \begin{cases} 1 & \lfloor M+1 \rfloor odd \\ 0 &\lfloor M+1 \rfloor even\end {cases}$

• 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.

• This question is odd as it leaves to the potential answerer the task to find out what happens with any possible candidate $\,L\,$ to be a limit...and with epsilon and delta! Can't you prove by showing two different sequences converging to infinity but for which the value of $\,x_n\sin x_n\,$ is different? – DonAntonio May 13 '13 at 17:06
• We were asked to solve the question both ways (sequences and delta epsilon) – jreing May 13 '13 at 17:08
• Thanks for the answers, but I was wondering if anyone could say why my solution wrong (or okay)? – jreing May 13 '13 at 17:46
• Take a sequence that goes to negative infinity and a sequence that goes to positive infinity – Matthew Levy Dec 21 '14 at 18:06

Assume the limit exists and is $L$. Fix $\epsilon =1$. There exists $M>0$ such that for $x>M$, $|x\sin{x}-L|<\epsilon=1$. Set $P=max\{L,M\}$.

Let $x_{0}=\left(\frac{\pi}{2}+2P\pi\right)$. We have, $x_{0}>M$ and $x_{0}\sin{x_{0}}=\left(\frac{\pi}{2}+2P\pi\right)$. Then $$|x_{0}\sin{x_{0}}-L|>1$$ Hence we have a contradiction.

Solution by means of sequences:

$$x_n:=\pi n\implies x_n\sin x_n=0\xrightarrow[n\to\infty]{}0$$

$$y_n:=\frac{(4n-3)\pi}2\implies y_n\sin y_n=\frac{(4n-3)\pi}2\xrightarrow[n\to\infty]{}\infty$$

First, you should convince yourself with a picture:

Second, you need to show that for any $L$, that there exists an $\epsilon>0$ such that for any $N$, there exists some $x\ge N$ such that $|L-x \sin x| \ge \epsilon$.

In this example, one easy way to show this is to let $x_n = (\frac{1}{2}+2n)\pi$, and note that $x_n \sin x_n = x_n$. So pick any $L$, and let $\epsilon=1$. Now choose $N$ and select $k$ such that $x_k > \min(L+1, N)$. Then $|L-x_k \sin x_k| > 1$.

Proof: Since the value of $f(x) = x\sin(x)$ oscillate in interval $[-x, x]$, then for arbitrary number $N$, I am always able to choose $~x_{1},x_{2}\gt N$, such that $f(x_{1})=0$ and $f(x_{2})=1$. Assume the limit exist as $x$ goes to infinity, then

$$\forall\varepsilon\gt0,\exists N\gt0,\text{such that } x \gt N,~|f(x)-L| \lt \varepsilon \tag{1}$$

Let $\varepsilon = \frac{1}{2}$. Because $~x_{1},x_{2}\gt N$，then by $(1)$, we must both have: \begin{align*} \ &|f(x_{1}) - L|= |L| \lt \frac{1}{2} \tag{2} \\&|f(x_{2})-L|=|1-L|\lt \frac{1}{2} \tag{3} \end{align*}

(2),(3) inequality cannot simultaneously be true for L. Thus by contradiction, there is no limit of $x\sin(x)$ as $x$ towards infinity.

you can always choose $x=k\pi$, then you can show by the $\epsilon-\delta$ language that if the limit exists, it must be $0$. But then, you choose $x=2k\pi+\frac\pi2$, which shows that the limit might not be $0$. Ok, I have used the uniqueness of limits in $R$, but somehow you can always choose some special $x$, which makes $sinx$ become a constant to solve this question.