Determine whether or not the limit exists: $\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2}$

Consider $$f(x,y)=\frac{(x+y)^2}{x^2+y^2} .$$ If you take the path $(0,y)$, then: $$\displaystyle\lim_{y\to0} f(0,y) =\lim_{y\to0} \frac{y^2}{y^2}=1$$

If you take the path $(x,x)$, then: $$ \lim_{x\to0}f(x,x) =\lim_{x\to0} \frac{(2x)^2}{2x^2}=2$$

So, the limit doesn't exist.

Although the question has been answered aptly, it might be instructive to see the application of polar coordinate transformation in the analysis of limits of the type herein.

To that end, we transform the problem to polar coordinates. Letting $x=\rho \cos\phi$ and $y=\rho \sin \phi$, we have

$$\frac{(x+y)^2}{x^2+y^2}=1+\sin 2\phi$$

Obviously, the limit

$$\lim_{\rho \to 0} (1+\sin 2\phi)$$

is highly dependent on the migration of $\phi$ as $\rho \to 0$. Thus, the limit does not exist.

Previous answers have as paths $1)$ $\phi =\pi/2$ for which the limit is $1$ and $2)$ $\phi = \pi/4$ for which the limit is $2$.

Here's another way $$ \lim\limits_{(x,y)\to(0,0)} \frac{(x+y)^2}{x^2+y^2} $$ Using polar coordinates, we have $$ \lim\limits_{r\to 0^+} \frac{\left(r\cos\phi+r\sin\phi\right)^2}{r^2\cos^2\phi + r^2\sin^2\phi} $$ $$ = \lim\limits_{r\to 0^+} \frac{r^2\left(\cos\phi+\sin\phi\right)^2}{r^2\left(\cos^2\phi + \sin^2\phi\right)} $$ $$ = \lim\limits_{r\to 0^+} \frac{\left(\cos\phi+\sin\phi\right)^2}{\cos^2\phi + \sin^2\phi} $$ $$ = \lim\limits_{r\to 0^+} \left(\cos\phi+\sin\phi\right)^2 $$ $$ = \sin(2\phi)+1 $$ This limit is clearly dependent on $\phi$. Therefore $$ \lim\limits_{(x,y)\to(0,0)} \frac{(x+y)^2}{x^2+y^2}\ \mbox{does not exist} $$