Limit of a two variable easy function

Question

I've been trying to solve this limit using polars and I get that it does not exist. But wolfram alpha gives a result of 1. $$ \lim_{(x,y) \to (0,0)} \frac{(x-y)^2}{(x^2+y^2)}$$

Answer 1

Hint

Taking the limit along the straight lines $y=mx$ will show that these limits depend on $m$, so...

$$\lim_{(x,y) \to (0,0)} \frac{(x-y)^2}{x^2+y^2}\xrightarrow{y=mx}\lim_{x \to 0} \frac{(x-mx)^2}{x^2+m^2x^2}=\frac{(1-m)^2}{1+m^2}$$

But wolfram alpha gives a result of 1.

I get (link): "(limit does not exist, is path dependent, or cannot be determined)".

Answer 2

Note that the limit does not exist, indeed

• for $x=y=t$

$$\lim_{(x,y) \to (0,0)} \frac{(x-y)^2}{(x^2+y^2)}=\lim_{t\to0} \frac{(0)^2}{2t^2}=0$$

• for $x=-y=t$

$$\lim_{(x,y) \to (0,0)} \frac{(x-y)^2}{(x^2+y^2)}=\lim_{ t\to0} \frac{(2t)^2}{2t^2}=\lim_{ t\to0} \frac{4t^2}{2t^2}=2$$

Answer 3

Using polar coordinates, which are:

$$\begin{cases} x=r\cos \theta \\ y=r\sin \theta\end{cases}$$

you get for $r>0$:

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

This does not have a limit as $r\to 0$, since it will give you many different values for different angles $\theta$.

Wolfram must be mistaken.

Answer 4

Easy : Let $y=0$ then $\lim_{x\to0}\frac{x^2}{x^2}=1$ And : put $x=y$ we get $\lim_{(x,y)\to(0,0)}\frac{(x-x)^2}{2x^2}=0$ So the answer no limit