Find $\lim_{(x,y) \to (0,0)} \frac{xy-\sin(x)\sin(y)}{x^2+y^2}$

Question

Find $$\lim_{(x,y) \to (0,0)} \frac{xy-\sin(x)\sin(y)}{x^2+y^2}$$

What I did was:

• Find the second order Taylor expansion for $xy-\sin(x)\sin(y)$ at $(0,0)$:

$P_2 (0,0)=f(0,0)+df_{(0,0)}+d^2f_{(0,0)}+r_2(0,0)$

1. $f(0,0)=0$
2. $df_{(0,0)}=\frac{\partial f}{\partial x} (0,0) \Delta x + \frac{\partial f}{\partial y} (0,0) \Delta y = 0$
3. $d^2f_{(0,0)}=\frac{\partial f}{\partial x} (0,0) \Delta x^2 + \frac{\partial f}{\partial x \partial y} (0,0) \Delta x \Delta y + \frac{\partial f}{\partial y} (0,0) \Delta y^2 = 0$

$\implies$ $P_2(0,0)=r_2(0,0)$

• Now I have $\lim_{(x,y) \to (0,0)} \frac{r_2(0,0)}{x^2+y^2}$

$\implies \lim_{(x,y) \to (0,0)} \frac{r_2(0,0)}{x^2+y^2} = \lim_{(x,y) \to (0,0)} \frac{r_2(0,0)}{x^2+y^2} \cdot \frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}} = \lim_{(x,y) \to (0,0)} \frac{r_2(0,0)}{\sqrt{x^2+y^2}} \cdot \frac{\sqrt{x^2+y^2}}{x^2+y^2}= 0$

However, wolframalpha is telling me the limit does not exist...

Answer 1

We have that

$$\frac{xy-\sin(x)\sin(y)}{x^2+y^2}=\frac{xy}{x^2+y^2}\left(1-\frac{\sin(x)}x\frac{\sin(y)}y\right)\to 0$$

indeed $\frac{xy}{x^2+y^2}$ is bounded and $1-\frac{\sin(x)}x\frac{\sin(y)}y \to 0$

As an alternative by polar coordinates

$$\frac{xy-\sin(x)\sin(y)}{x^2+y^2}=\cos\theta\sin \theta-\frac{\sin(r\cos\theta)}r\frac{\sin(r\sin\theta)}r=$$

$$=\cos\theta\sin \theta-\left(\cos \theta+o(r)\right)\left(\sin \theta+o(r)\right)=o(r) \to 0$$

Answer 2

You can squeeze the expression as follows using

• $x^2+y^2 \geq 2|xy|$

Consider only $xy\neq 0$, otherwise there is nothing left to show:

$$0\leq \frac{|xy-\sin x \sin y|}{x^2+y^2}\leq \frac{|xy-\sin x \sin y|}{2|xy|}=\frac 12\left|1-\frac{\sin x \sin y}{xy} \right|\stackrel{(x,y)\to (0,0)}{\longrightarrow}0$$