A limit that involves two variables

Question

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?

Answer 1

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

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 $\sqrt{(x-0)^2+(y-0)^2} < \delta$

then

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

Answer 2

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

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