# Does the limit exist or is wolframalpha right?

I am looking to solve this equation: $$\lim_{(x,y)\to(0,0)} \frac{y\sin(x-y)}{(x^2+y^2)^{5/8}}$$

According to wolframalpha, the limit does not exist.

I tried finding different paths that we can receive different values, but couldn't. So I thought how about trying to prove it and seeing where I fail and trying to find a path then.

Using sandwish, I did the following:

$$0\le\left|\frac{y\sin(x-y)}{(x^2+y^2)^{5/8}}\right|\le\left|\frac{|y||x-y|}{(x^2+y^2)^{5/8}}\right|\le\left|\frac{|yx-y^2|}{(x^2+y^2)^{5/8}}\right|\le\left|\frac{|y^2|+|xy|}{(x^2+y^2)^{5/8}}\right|$$

And here is how I see it doesn't exist (although I am not sure), but how do I proof it doens't? I can't find a path that doesn't exist.

I tried all paths, like $$x=y, x=y^{1/2}, x=y^2, x=y^3$$, and in all of them, the limit is zero. Therefore I tend to believe it exists, yet I can not solve it.

• Yes, my bad; it was added by mistake, I'll fix it and add <= – Rab Jul 9 '17 at 14:16
• The $\sin$ is making WolframAlpha give up. Observe that $x-y \to_{(x,y)\to 0} 0$, and using $\frac{\sin u}{u}\to_{u\to 0}1$ ask for $$\lim_{(x,y)\to(0,0)} \frac{y(x-y)}{(x^2+y^2)^{5/8}}$$ instead. – Clement C. Jul 9 '17 at 14:35

## 3 Answers

The $\sin$ is making WolframAlpha fail. Observe that $\frac{\sin u}{u}\xrightarrow[u\to 0]{}1$ and use this observation to ask for the limit of $$\left\lvert \frac{y(x-y)}{(x^2+y^2)^{5/8}}\right\rvert$$ instead; you will have more luck.

Now, to actually prove the limit is $0$:

Observe that $$\frac{1}{\sqrt{2}}\lVert (x,y)\rVert_1\leq \sqrt{x^2+y^2} = \lVert (x,y)\rVert_2 \leq \lVert (x,y)\rVert_1$$ (one side by Cauchy—Schwarz, the other by monotonicity; or, if you don't care about constants, "all norms are equivalent in finite dimension.")

Thus, you have that \begin{align}0 \leq \left\lvert \frac{y\sin(x-y)}{(x^2+y^2)^{5/8}}\right\rvert &\leq \left\lvert \frac{y(x-y)}{\lVert (x,y)\rVert_2^{5/4}}\right\rvert \leq \frac{\lvert y(x-y)\rvert }{\lVert (x,y)\rVert_2^{5/4}} \leq \sqrt{2} \frac{\lvert y(x-y)\rvert }{\lVert (x,y)\rVert_1^{5/4}} = \sqrt{2} \frac{\lvert y\rvert (\lvert x\rvert+\lvert y\rvert) }{(\lvert x\rvert+\lvert y\rvert)^{5/4}}\\ &\leq \sqrt{2} \frac{\lvert y\rvert}{(\lvert x\rvert+\lvert y\rvert)^{1/4}}\leq \sqrt{2} \frac{\lvert x\rvert+\lvert y\rvert}{(\lvert x\rvert+\lvert y\rvert)^{1/4}} = \sqrt{2} (\lvert x\rvert+\lvert y\rvert)^{3/4} \xrightarrow[(x,y)\to 0]{}0\end{align} and you can conclude.

If $x$ and $y$ are real, the limit is $0$: Your estimate shows $$\left|\frac{y \sin(x - y)}{(x^{2} + y^{2})^{5/8}}\right| \leq \frac{(|x| + |y|)^{2}}{(x^{2} + y^{2})^{5/8}} \leq 4(x^{2} + y^{2})^{3/8}.$$ (For the second inequality, write $(x, y)$ in polar coordinates, and note that $(|\cos\theta| + |\sin\theta|)^{2} \leq 4$ for all real $\theta$.)

Wolfram Alpha states the limit does not exist "in complex space"; if $x = -iy \neq 0$, for example, the numerator does not vanish but the denominator does. That is, the expression isn't even defined throughout a neighborhood of the origin, nor is the expression bounded in magnitude "close to" the origin.

hint

use $$x=r\cos (t) \;\;, \;y=r\sin (t)$$

and

$$|\sin (x-y)|\le r|\cos (t)-\sin (t)|\le 2r$$