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WolframAlpha says $$ \lim_{x\to 0}\biggl(\lim_{y\to 0} x\sin\frac{1}{y}\biggl) $$ does not exists.

But I think when $y\to0$, $x\sin\frac{1}{y}$ tends to $\{-x,x\}$, so when $x\to0$ the limit is $\{0,0\}$

So what exactly the limit is? Does it exist?

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    $\begingroup$ Are you aware that $\lim_{y\to 0}\lim_{x\to 0}$ is not the same as $\lim_{(x,y)\to (0,0)}$? Moreover, note that $\lim_{y\to 0 }\sin(1/y)$ does not exists. $\endgroup$
    – Surb
    Apr 1, 2015 at 15:02

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The "iterated" limit $$ \lim_{x\to 0} \lim_{y\to 0} x\sin \Bigl(\frac{1}{y}\Bigr)$$ does not exists, because it does not exist the inner limit.

EDIT: my first treatment of the limit for $(x,y)\to (0,0)$ was not correct. I think that now all works well.

We have to be careful about the following limit in $\mathbb{R}^2$ $$\lim_{(x,y)\to (0,0)} x\sin \Bigl(\frac{1}{y}\Bigr)$$ Call for simplicity $f(x,y) = x\sin \Bigl( \frac{1}{y}\Bigr)$.

The point $(0,0)$ is an accumulation point for pairs of the form $(x,0)$ where the inner function is not defined. It turns out that the limit does not exist, because the sequence $(\frac{1}{n},0)$ converges to $(0,0)$ as $n$ approaches $+\infty$, but $f(\frac{1}{n},0)$ does not converge to any value (since $f$ is not defined at every point of the sequence); while at the same time, there are infinitely many sequences $(x_n,y_n)$ converging to $(0,0)$ such that $f(x_n,y_n)$ converges to $0$.

However, it is meaningful to consider the limit $$\lim_{\substack{(x,y)\to (0,0) \\ y \neq 0}} x\sin \Bigl(\frac{1}{y}\Bigr)$$ We are simply restricting ourselves to the domain of $f$. In this region, which is $D = \mathbb{R}^2 \setminus \{y=0\}$, the inner function is everywhere defined and we can argue in a simple way: the sine is bounded everywhere and the function $x$ is infinitesimal as $(x,y)\to (0,0)$. Since this estimate holds in the whole $D$, we conclude that the above limit exists and is equal to $0$.

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  • $\begingroup$ I think $$\lim_{(x,y)\to (0,0)} x\sin \Bigl(\frac{1}{y}\Bigr)$$ exists and equals to 0.You can varify in the wolframalpha $\endgroup$
    – partida
    Apr 1, 2015 at 15:40
  • $\begingroup$ @user15961, did you check all trajectories? $\endgroup$ Apr 1, 2015 at 15:42
  • $\begingroup$ @jameselmore sin(1/y) have boundry,however x is an Infinitesimal. Infinitesimal times boundry value is Infinitesimal,isn't it? $\endgroup$
    – partida
    Apr 1, 2015 at 15:48
  • $\begingroup$ @user15961, if we approach this limit along the trajectory $(x,y) = \lim_{t\to 0} (t,0)$, then this limit is undefined. Because of this we can't claim (in general) that the $\lim_{(x,y)\to (0,0)}$ is defined $\endgroup$ Apr 1, 2015 at 15:54
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We've been asked to consider $\lim_{x \to 0}\big(\lim_{y\to 0} x\cdot\sin(\frac1y)\big)$. Which is undefined due to the 'iterative' nature of it's definition. In response, people have made several claims which have considered the same limit under arbitrary trajectories $(x,y) \to (0,0)$. Let's look at this a little closer. If we consider $r,\theta$ such that $x=r\cos(\theta), y = r\sin(\theta)$

$$L = \lim_{(x,y)\to (0,0)}x\cdot\sin(\frac1y) = \lim_{r\to0} r\cos(\theta)\cdot \sin(\frac{1}{r\sin(\theta)})$$ If we assume for the moment that $\sin(\theta) \neq 0$, then $$0 = \lim_{r\to 0} -r|\cos(\theta)|\leq L \leq \lim_{r\to 0} r|\cos(\theta)| = 0$$ So it appears that $$L = 0 \ \forall\theta\neq n \pi, n\in\mathbb{N},\text{but }L\text{ DNE, elsewhere.}$$

Because of this, under certain trajectories this limit can tend to $0$ but since it is not $0$ under all trajectories (and we have only looked at constant vector trajectories), we cannot conclude that this limit exists.

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If I understand correctly, with writing:

limit(limit(x*sin(1/y),y->0),x->0)

you asked Wolfram Alpha to calculate first the limit $$\lim_{y\rightarrow 0}$$ In this way you are considering x as a parameter. For this Wolfram Alpha says that the limit does not exist!

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