I'm trying to compute this limit $\lim_{(x,y) \to (0,0)}2x\sin^2(\frac{1}{y})$, but WolframAlpha says that it does not exist.

I'm not quite sure why. I do understand that there are oscillations coming from the $\sin(1/y)$. However, $x \to 0$ as well. Shouldn't that crush the function to zero?

Also, I know that $\lim_{x \to 0} x \sin(\frac{1}{x}) = 0$. Isn't that pretty much the same idea as the limit in question?

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    $\begingroup$ The function isn't defined on the $x$-axis. If you don't worry about that, then $|f(x,y)|\le 2|x|$ and so tends to zero, as you say. With Wolfie A you get what you pay for. $\endgroup$ – Lord Shark the Unknown Feb 2 '18 at 6:54
  • $\begingroup$ Why is it not defined on the x-axis? $\endgroup$ – user1691278 Feb 2 '18 at 7:00
  • $\begingroup$ @LordSharktheUnknown I see what you mean. How can I not worry about that though? $\endgroup$ – user1691278 Feb 2 '18 at 7:05
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    $\begingroup$ @user1691278 No, that would be defined. The line that lies exactly on the $x$ axis, and only that line is undefined. So if we force $y=0$, things don't work. If you don't care about that particular path, then the limit does exist. i.e the limit is $0$ for every path except the one on the $x$ axis. $\endgroup$ – Mitchell Faas Feb 2 '18 at 7:13
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    $\begingroup$ @user1691278 In the 1-D case this is indeed what happens, but try to think of it in terms of distances and points. In 1-D space: Suppose we pick a point $p$ ($0$ in this case) and a variable $x$. We wish to compute the limit when the distance $d(p,x)$ between $p$ and $x$ tends to $0$ (but does not actually attain it). In the 2-D space we do the same: We pick a point $p, ((0,0))$ in this case) and compute the limit as $d((0,0),(x,y))$ tends to $0$. But if we choose $x=1, y=0$ we still have a non-zero distance: $d((0,0), (1,0)) = 1$. So we can force $y=0$ and still compute the limit. $\endgroup$ – Mitchell Faas Feb 2 '18 at 7:24

It is equal to $0$ according to Wolfram alpha.

enter image description here

The domain of the function should exclude the $x$-axis, that is the domain $D = \{ (x,y) : y \neq 0\}$.

Let $\epsilon > 0$, we choose $\delta = \frac{\epsilon}2$, if $(x,y) \in D$ and $\sqrt{(x-0)^2+(y-0)^2} < \delta$


$$ \left|2x\sin^2 \left( \frac1y\right)-0\right| = \left|2x\sin^2 \left( \frac1y\right)\right|\leq 2|x| \leq 2 \delta < \epsilon.$$

  • $\begingroup$ I don't have the pro version, but I see this: wolframalpha.com/input/… $\endgroup$ – user1691278 Feb 2 '18 at 7:01
  • $\begingroup$ that's interesting thing that wolfram alpha gives different answer according to versions. $\endgroup$ – Siong Thye Goh Feb 2 '18 at 7:02
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    $\begingroup$ I presume one has to pay for the Pro version. $\endgroup$ – Lord Shark the Unknown Feb 2 '18 at 7:05
  • $\begingroup$ @LordSharktheUnknown How can you have two answers though? $\endgroup$ – user1691278 Feb 2 '18 at 7:06
  • $\begingroup$ @SiongThyeGoh The problem is that with the non-pro version, the allotted computational time gets exceeded. So wolfram just says it doesn't exist, adding that it may be dependent upon the path in complex space. i.e It hasn't been able to find the answer in a small enough time frame. $\endgroup$ – Mitchell Faas Feb 2 '18 at 7:10

$$|2x\sin^2(1/y)|\le 2|x|\cdot |1|\le 2|x|.$$

Since $2|x|\to 0$ as $x\to 0$, the limit is not undefined.

  • $\begingroup$ Doesn't this contradict @MitchellFaas's comment? $\endgroup$ – user1691278 Feb 2 '18 at 14:30
  • $\begingroup$ @user1691278 idk what coment he made $\endgroup$ – Asim90 Feb 3 '18 at 5:25
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    $\begingroup$ @user1691278 It does not :) The limit is defined as long as $y\neq 0$. The inequality Asim90 writes no longer holds in this case: Say $y = 0$ Then $1/y$ is undefined, so $2x\sin^2(1/y)$ is undefined, so we do not know its relationship with the right part of the equality. This has to do with the domain of $\sin(x)$. Think of it like this: Does the inequality $|\sin(\{monkey\})|\leq1$ make any sense? If that doesn't, then surely $|\sin(\{Undefined\})|\leq1$ makes no sense. $\endgroup$ – Mitchell Faas Feb 3 '18 at 18:55

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