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 $(0,0)$:
$P_2 (0,0)=f(0,0)+df_{(0,0)}+d^2f_{(0,0)}+r_2(0,0)$
- $f(0,0)=0$
- $df_{(0,0)}=\frac{\partial f}{\partial x} (0,0) \Delta x + \frac{\partial f}{\partial y} (0,0) \Delta y = 0$
- $d^2f_{(0,0)}=\frac{\partial f}{\partial x} (0,0) \Delta x^2 + \frac{\partial f}{\partial x \partial y} (0,0) \Delta x \Delta y + \frac{\partial f}{\partial y} (0,0) \Delta y^2 = 0$
$\implies$ $P_2(0,0)=r_2(0,0)$
- 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…
Best Answer
You can squeeze the expression as follows using
Consider only $xy\neq 0$, otherwise there is nothing left to show:
$$0\leq \frac{|xy-\sin x \sin y|}{x^2+y^2}\leq \frac{|xy-\sin x \sin y|}{2|xy|}=\frac 12\left|1-\frac{\sin x \sin y}{xy} \right|\stackrel{(x,y)\to (0,0)}{\longrightarrow}0$$