Does $\lim_{(x,y)\rightarrow (0,0)}\frac{(x^2-y^2) \cos (x+y)}{x^2+y^2}$ exists? 

Does
\[\lim_{(x,y)\rightarrow (0,0)}\frac{(x^2-y^2) \cos (x+y)}{x^2+y^2}\]
exists?

Answer

Writing the limit in polar coordinates (i.e. $x=r \cos \theta$ and $y=r \sin \theta$) we get

\[\lim_{(x,y) \rightarrow (0,0)}\frac{(x^2-y^2)\cos(x+y)}{x^2+y^2}=\lim_{r \rightarrow 0} \frac{r^2(\cos^2 \theta-\sin^2 \theta) \cos(r(\cos \theta +\sin \theta))}{r^2}\] 
\[=\lim_{r \rightarrow 0} (\cos^2 \theta-\sin^2 \theta) \cos(r(\cos \theta +\sin \theta))\]
\[=\cos^2 \theta-\sin^2 \theta. \]
Since the above depends on $\theta$ the limit is question does not exists. Indeed on the line $y=x$ we have $\theta=\frac{\pi}{4}$ 
\[\cos^2 \theta-\sin^2 \theta =(\frac{\sqrt{2}}{2})^2-(\frac{\sqrt{2}}{2})^2=0.\]
On the line $y=0$ we have $\theta=0$, and hence 
\[\cos^2 \theta-\sin^2 \theta =1^2-0^2=1.\]
Thus the limit does not exist, since if it exists it must be unique. 

The answer is accepted.