Evaluate $\lim\limits_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2}$

Question

Evaluate $\displaystyle\lim_{(x,y)\to(0,0)}\dfrac{(x+y)^2}{x^2+y^2}$

Using polar, we have $x=r\cos(\theta),y=r\sin(\theta)$

Our limit becomes:

$$\lim_{r\to 0}\dfrac{(r\cos(\theta)+r\sin(\theta))^2}{r^2\sin^2(\theta)+r^2\cos(\theta)}=\lim_{r\to 0}\dfrac{r^2\cos^2(\theta)+2r^2\cos(\theta)\sin(\theta)+r^2\sin^2(\theta)}{r^2}$$

Factoring and dividing removes the $r^2$ in the denominator, and we get $1$ as the limit.

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However this is not right.

If we consider along the $x-axis$, our limit becomes $1$. If we consider along the line $y=x$, our limit becomes $1/2$, and are clearly not equal.

This means that the limit does not exist but my polar said it does and it equals $1$. Where did I mess up in my polar coordinates?

Answer 1

Its been extensively discussed and OP is probably a bit exhausted by seeing many view points that some are clear and some are not. Let's give a "nice proof " that the limit does not exist. Just choose two paths: path $1$ is: $x = -y = t$, then the limit is $0$, and path $2$ is: $x = y = t$, then the limit is $2$. Different values of limits show there is no limit at $(0,0)$.