Compute $\lim\limits_{(x,y) \to (0,0)} \frac{x^2}{x^2+y^2} \sin\left(\frac{xy}{\sqrt{x^2 + y^2}}\right)$

limitsmultivariable-calculus

Acordding to Wolfram Alpha:

$$\lim\limits_{(x,y) \to (0,0)} \frac{x^2}{x^2+y^2} \sin\left(\frac{xy}{\sqrt{x^2 + y^2}}\right) \quad\text{does not exist.}$$

Using:

$$\gamma(t) = (t,0)\;\; \text{and}\;\; \gamma(t) = (t,t)$$ we can easily prove that

$$\lim\limits_{(x,y) \to (0,0)} \frac{x^2}{x^2+y^2} \quad\text{does not exist.}$$

however

$$\lim\limits_{(x,y) \to (0,0)} \sin\left(\frac{xy}{\sqrt{x^2 + y^2}}\right) = 0$$

so there are no paths that will give a result different than $0$.

My question has two parts:

  1. How do I prove that this limit does not exist.
  2. On most of the online courses that I take, in the questions they explicitly tell you if the limit exists before you start. However, my professor in the exams just tells us to calculate the limit or prove if it does not exist. Is there a easy way to determine if the limit exists? I have expend around 30 minutes trying to prove that this limit exists by the squeeze theorem, before searching the answer on Wolfram Alpha.

Edit: People in the comments said that this limit does exist. If this is the case, my question becomes how to compute it.

Best Answer

From the triangle inequality we have

$$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\le \frac12 \sqrt{x^2+y^2}$$

Therefore, using $0\le \frac{x^2}{x^2+y^2}\le 1$ and $\left|\sin(\theta)\right|\le \left|\theta\right|$, we have

$$\left|\frac{x^2}{x^2+y^2}\sin\left(\frac{xy}{\sqrt{x^2+y^2}}\right)\right|\le \frac12\sqrt{x^2+y^2}$$

And you can finish now.

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