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I found this interesting problem on calculating the limit of $\frac{\sin(xy^2)}{xy}$ on the positive coordinate axes $x$ and $y$. That is, compute the limit on the points $(x_0, 0)$ and $(0,y_0)$ when $x_0 > 0$ and $y_0 > 0$.

My approach was this:

If we first calculate the limit for $x$ axis, the the $x$ is a constant $x=x_0$ and therefore the function is essentially a function of one variable:

$$f(y) = \frac{\sin(x_0y^2)}{x_0y}$$

Using L'Hospital's rule:

$$\lim_{y\to0}f(y)=\frac{\lim_{y\to0}\frac{d\sin(x_0y^2)}{dy}}{\lim_{y\to0}\frac{dx_0y}{dy}}$$

$$=\frac{\lim_{y\to0}2yx_0\cos(x_0y^2)}{\lim_{y\to0}x_0}=0$$

The same idea applies to the other limit.

But in the sheet where this problem was listed, this was listed as a "tricky" limit to calculate. It seemed quite simple, so I would like to hear whether this approach is correct. Thank you!

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It is correct, the limit is $0$ and the reasoning holds.

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  • $\begingroup$ Thank you for the answer! However, I thought occured to me: Haven't I just proven just that the limit is zero if we approach the axis along a normal line to the axis? But in two dimensions, limits often depend on the path we take to approach the point. So could it be possible there are some other paths (like quadratic) that give a different value? $\endgroup$ – S. Rotos Feb 17 at 20:54
  • $\begingroup$ True, but $f_{x_0}(y)$ is continuous on all directions of space, so you will find zero nonetheless. You need to calculate the limit with $x$ and you'll find 0 aswell. There exists functions that have different derivatives depending on the path you take. $\endgroup$ – PackSciences Feb 17 at 20:58
  • $\begingroup$ I remember reading somewhere that even though all directional derivatives exist on a point, it does not guarantee that the function has a limit at that point. Do I remember wrong or did I misunderstand something? $\endgroup$ – S. Rotos Feb 17 at 21:09
  • $\begingroup$ I don't think I have enough knowledge to answer this. Maybe someone else can. $\endgroup$ – PackSciences Feb 17 at 21:11
  • $\begingroup$ I found the mention, and I remembered I wrong, it actually said that the existence of partial derivatives is not enough. But what do you mean when you say $f_{x_0}$ is continuous in all directions? What partial derivative is this? $\endgroup$ – S. Rotos Feb 17 at 21:15
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$$\lim_{x,y\to0}\Big(\frac{\sin{(xy^2)}}{xy}\Big)=\lim_{x,y\to0}\Big(\frac{xy^2}{xy}\Big)$$ $$=\lim_{x,y\to0}y=0$$ The first line is based on the fact that $\sin{(x)} \approx x$ for small $x$ which can be proven by studying its Taylor series expansion.

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  • $\begingroup$ I think this only holds for the limit for the x axis. Note that the problem was not to take the limit to the origin, but to the positive coordinate axes. $\endgroup$ – S. Rotos Feb 17 at 23:01
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Yes, your reasoning is correct. The tricky part is what it implies! $$\lim_{(x,y) \to (x_0, 0)} \frac{\sin(xy^2)}{xy} = 0 \qquad \text{and} \qquad \lim_{(x,y) \to (0,y_0)} \frac{\sin(xy^2)}{xy} = 1$$ Now imagine taking limits as $x_0$ and $y_0$ go to 0. You're going to the same point $(0,0)$ along two different paths, and getting two different answers for what the limit should be!

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