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Find $$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (xy)}{x+y}$$ exist or DNE.

$f(x,y)$ along the lines $y=mx$ \begin{align} \lim_{(x,mx)\rightarrow (0,0)}\frac{\sin (mx^2)}{x+mx}&=\lim_{x\rightarrow 0}\left(\frac{\sin (mx^2)}{x}\frac{1}{1+m}\right)\\ &=0 \end{align} By applying L'Hospital's Rule, we can show this limit is $0$ except when $m=-1$ .
But the answer says limit DNE. Maybe I have to choose different path. Then I think whyn't choose $y=mx^n$ and again find limit $0$. I don't use polar or spherical coordinates because here is no $x^2+y^2$ terms. Eventually I can't use $\lim_{p\rightarrow a}f(g(p))=f(\lim_{p\rightarrow a}g(p))($usage of continuiuty$)$ also. Is there another approach for this problem$?$

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  • $\begingroup$ you can't split the limit like this. (the limit of the product is not equal in general to the product of the limit (see for instance $x/x$ when x goes to 0) $\endgroup$
    – Jeanba
    Dec 2, 2019 at 14:29
  • $\begingroup$ The function $f(x,y)=\frac{\sin(xy)}{x+y}$ is not defined on the line $y=-x$, one needs to be clear what $(x,y)\to(0,0)$ means in this case since $f$ is not defined for all points "near" $0$. $\endgroup$
    – user9464
    Dec 2, 2019 at 14:34
  • $\begingroup$ @Jack I edited my post $\endgroup$
    – emonHR
    Dec 2, 2019 at 14:37

4 Answers 4

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No, the limit doesn't exist. See what happens when $(x,y)$ is of the form $\left(-\frac1n+\frac1{n^2},\frac1n\right)$ ($n\in\mathbb N$).

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  • $\begingroup$ How did you come up with this form$?$ It seems very interesting to me. $\endgroup$
    – emonHR
    Dec 2, 2019 at 14:40
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    $\begingroup$ The donominator is $x+y$, right?! So, I chose sequence of points $\bigl((x_n,y_n)\bigr)_{n\in\mathbb N}$ converging to $(0,0)$ such that $(x_n+y_n)_{n\in\mathbb N}$ converges to $0$ even faster than that. It works very often. $\endgroup$ Dec 2, 2019 at 14:52
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Your argument fails because you are only exploring a prticular family of paths wich is not sufficent to show that the limit exists.

To proceed, we have that

$$\frac{\sin (xy)}{x+y}=\frac{\sin (xy)}{xy}\frac{xy}{x+y}$$

and $\frac{\sin (xy)}{xy} \to 1$ but $\frac{xy}{x+y}$ has no limit indeed

  • for $x=0 $

    $$\frac{xy}{x+y}=0$$

  • for $x=t$ and $y=-t+t^2$ with $t\to 0$

    $$\frac{xy}{x+y}=\frac{-t^2+t^3}{t-t+t^2}=-1+t \to -1$$


Edit

Unfortunately there are not general rules to find critical paths and we need to proceed case by case.

In that case it easy to find the path for which the limit is equal to $0$.

For the other path, a good strategy which often works, is to select at first a path such that the denominator is equal to zero that is $x=-y$ in this case but since $f(x,y)$ is not defined at those points we add to $y$ an extra smaller term that is $t^2$. In some cases the first guess doesn't work and we need to use a different extra term for $y$.

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  • $\begingroup$ How did you guess $x=t$ and $y=-t+t^2?$ Because it seems not working for $y=x^2$ and $y=mx$. I really appreciate every answer but without explaining it make no sense to me @user $\endgroup$
    – emonHR
    Dec 2, 2019 at 14:49
  • $\begingroup$ @emonHR Yes of course! The key point is to explore the paths such that $x+y$ vanishes. Of course we can't take $x=t$ and $y=-t$ therefore we try with something with has some "small" remainder. I add something on that. $\endgroup$
    – user
    Dec 2, 2019 at 14:52
  • $\begingroup$ thanks for explaining(+1) @user $\endgroup$
    – emonHR
    Dec 2, 2019 at 14:58
  • $\begingroup$ @emonHR You are welcome! Take a look also to this other example $\endgroup$
    – user
    Dec 2, 2019 at 15:01
  • $\begingroup$ suddenly I think why not choose $y=\sin x?$ and trying further I got $y=-\sin x$ gives $\frac{2}{0}$. which mean limit DNE. Can I post my answer $?$ @user $\endgroup$
    – emonHR
    Dec 2, 2019 at 15:06
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HINT Your argument fails when $m=-1$.

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  • $\begingroup$ I think it also fails when $m=0$ $\endgroup$
    – Vasili
    Dec 2, 2019 at 14:34
  • $\begingroup$ @gt6989b I edited my post. $\endgroup$
    – emonHR
    Dec 2, 2019 at 14:36
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Hint

For $xy \ne 0$, rewrite $$\frac{\sin (xy)}{x+y} =\frac{\sin (xy)}{xy}\frac{xy}{x+y} $$ and since $$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (xy)}{xy}=1$$ the problem reduces to studying $$\lim_{(x,y)\rightarrow (0,0)}\frac{xy}{x+y}$$

Detailed information about this limit can be found at Does $\lim \frac{xy}{x+y}$ exist at (0,0)?

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