Prove $\lim_{(x,y) \to (0,0)} \frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2} =0$.

Question

I want to show that

$$\lim_{(x,y) \to (0,0)} \frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2} =0.$$

Is it valid to do it like this:

$$\lim_{(x,y) \to (0,0)} \left|\frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2}\right| \leq \lim_{(x,y) \to (0,0)} |\exp(xy)\cdot xy\cdot (x^2-y^2)|=0$$

Answer 1

We have that $e^{xy}\to 1$ and by polar coordinates

$$\left|\frac{xy\cdot(x^2-y^2)}{x^2+y^2}\right|=r^2|(\cos \theta\sin \theta)(\cos^2\theta-\sin^2\theta)|\le r^2\to 0$$

Answer 2

No, that's not valid. We have a/b 0 and b>1, but in this case we have the denominator going to less than 1, so that doesn't work. The $x^2+y^2$ suggests polar coordinates; that gives $$\frac{e^{r^2\sin(\theta)\cos(\theta)}r^2\sin(\theta)\cos(\theta)r^2(\cos^2(\theta)-\sin^2(\theta))}{r^2}=$$$$e^{r^2sin(\theta)cos(\theta)}r^2sin(\theta)cos(\theta)(cos^2(\theta)-sin^2(\theta))=$$$$\frac{e^{r^2sin(\theta)cos(\theta)}r^2sin(2\theta)cos(2\theta)}2=$$$$\frac{e^{r^2sin(2\theta)/2}r^2sin(4\theta)}4 $$

Answer 3

We may restrict the neibourhood to $x^2+y^2<1$, then $e^{xy}<2$ and $$\left|\dfrac{e^{xy}xy(x^2-y^2)}{x^2+y^2}\right|\leqslant\dfrac{2|xy|(|x|^2+|y|^2)}{x^2+y^2}\leqslant x^2+y^2$$