How to prove this limit does not exist?

Question

I have the following limit:

$$\lim_{(x,y) \to (0,0)} \frac{\sin(xy)}{\sqrt{x^2 + y^2 + |y|} \times \|(x,y)\|}$$

Where I'm using the usual euclidean norm. I've been trying to find a curve that approximates (0,0) in a way such that the limit is not 0, but I'm really struggling to do so.

Answer 1

We have that

$$\lim_{(x,y) \to (0,0)} \frac{\sin(xy)}{\sqrt{x^2 + y^2 + |y|} \times ||(x,y)||}=\lim_{(x,y) \to (0,0)} \frac{\sin(xy)}{xy} \frac{xy}{\sqrt{x^2 + y^2 + |y|} \times ||(x,y)||}$$

and

$$\lim_{(x,y) \to (0,0)} \frac{\sin(xy)}{xy}=1$$

and by polar coordinates

$$\frac{xy}{\sqrt{x^2 + y^2 + |y|} \times ||(x,y)||}=\frac{r\cos \theta \sin \theta}{\sqrt{r^2+r|\sin \theta|}}\to 0$$

indeed

$$0\le \frac{r|\cos \theta| |\sin \theta|}{\sqrt{r^2+r|\sin \theta|}}\le \frac{r|\cos \theta| |\sin \theta|}{\sqrt{r|\sin \theta|}} \frac{\sqrt{r|\sin \theta|}}{\sqrt{r|\sin \theta|}}=|\cos \theta|\sqrt{r|\sin \theta}| \to 0$$

Answer 2

I might be wrong, but I got that the limit does exist.

Multiply the equation with $(xy)/(xy)$ to simplify the limit to (using $\sin(xy)\approx xy$ if $xy$ close to $0$) $$ \lim_{(x,y) \to (0,0)} \frac{xy}{\sqrt{x^2 + y^2 + |y|} \cdot ||(x,y)||} $$ Write out $$ \sqrt{x^2+y^2+|y|}\cdot ||(x,y)|| = \sqrt{(x^2+y^2+|y|)(x^2+y^2)} = \sqrt{x^4 + 2x^2y^2 + y^4 + x^2|y| + y^2|y|} $$ If you divide this by $xy = \pm\sqrt{x^2y^2}$ (so, we are turning it around), you get $$ \pm\sqrt{\frac{x^2}{y^2} + 2 + \frac{y^2}{x^2} + \frac{1}{|y|} + \frac{|y|}{x^2}} $$ If you let $x$ and $y$ tend to $0$, this will always tend to $\pm\infty$ by the $+1/|y|$ term. Hence, if you revert the fraction it converges to $\pm0=0$. Thus the limit does exist and it should be $0$.

Answer 3

Using $|\sin(xy)|\le|xy|$ and then switching to polar coordinates, we have, for the square of the function,

$$\begin{align} {\sin^2(xy)\over(x^2+y^2+|y|)(x^2+y^2)} &\le{x^2y^2\over(x^2+y^2+|y|)(x^2+y^2)}\\ &={r\cos^2\theta\sin^2\theta\over r+|\sin\theta|}\\ &\le{r\sin^2\theta\cos^2\theta\over\max(r,|\sin\theta|)}\\ &=\min(\sin^2\theta\cos^2\theta,r|\sin\theta|\cos^2\theta)\\ &\le r|\sin\theta|\cos^2\theta\\ &\le r \end{align}$$

The squeeze theorem now tells us the limit of the original function is $0$, since its square tends to $0$ as $r\to0$.