Limits in multivariable

Question

$$\lim_{(x,y)\to (0,1)} \frac{x^2(y-1)^2}{x^2 + (y-1)^2}$$ What I did was: $$x = k(y-1)^2$$ Then: $$\lim_{(x,y)\to (0,1)}\frac{k(y-1)^2(y-1)^2}{k(y-1)^2 + (y-1)^2} = \lim_{(x,y)\to (0,1)}\frac{k(y-1)^2}{2} = 0$$

So is this a proof that the limit is defined and that it is 0? Or am I wrong and I need to do other things? In similar question they managed to lose all the variables and were left with only $k$ and they could then draw the conclusion that since k can be whatever value, the limit doesn't exist. But can I draw the same conclusion? I personally don't think I can't draw the same conclusion here because in my case I'm arguing that that limit exists but $x=k(y-1)^2$ is not the only way we can near that point. So what so I do to now?

Answer 1

A nice (in my opinion) aproach: define $\;h:=y-1\;$ so that $\;y\to 1\implies h\to 0\;$ , and thus your limit becomes

$$\lim_{(x,h)\to(0,0)}\frac{x^2h^2}{x^2+h^2}$$

but then:

$$\frac{x^2h^2}{x^2+h^2}=x^2\frac{h^2}{x^2+h^2}\le x^2\frac{h^2}{h^2}=x^2\cdot1\xrightarrow[(x,h)\to(0,0)]{}0$$

and we're done

Answer 2

Take $x=r\cos t$ and $y=r\sin t +1$. Now $$\lim_{(x,y)\to (0,1)} \frac{x^2(y-1)^2}{x^2 + (y-1)^2}=\lim_{r\to 0}\frac{r^4\sin^2 t\cos^2 t}{r^2}=0.$$ That proves that the limit exist and is equal to 0, your method is not enough since you would need to do the dame for every direction in $\mathbb{R}^2$.