3
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$\lim \limits_{(x,y) \to (0,1)} \frac{x^2(y-1)^2}{x^2+(y-1)^2}$

trying all lines $y=mx+1$ and $x=0$ yields $0$

So, let's try this $0$ as a candidate of L.

$$|\frac{x^2(y-1)^2}{x^2+(y-1)^2}-0| = \frac{x^2(y-1)^2}{x^2+(y-1)^2} \leq \frac{x^2(y-1)^2}{x^2} = (y-1)^2$$

which goes to $0$ as $y \rightarrow 1$

Hence by the squeeze/sandwich theorem

$$\lim \limits_{(x,y) \to (0,1)} \frac{x^2(y-1)^2}{x^2+(y-1)^2} = 0$$

Would this be a complete answer?

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  • $\begingroup$ Yes, it is correct. You just have to notice that $\frac{x^2(y-1)^2}{x^2+(1-y)^2}\leq (1-y)^2$ for all $(x,y)$ different of $(0,1)$. In your deduction you are assuming $x\neq 0$ because you are dividing by $x^2$. $\endgroup$ – Hugo C Botós Sep 20 '17 at 3:58
  • $\begingroup$ Yes because we don't actually care what happens at (0,1) right? We just want to know what happens as x approaches 0. $\endgroup$ – bolt997 Sep 20 '17 at 4:03
  • $\begingroup$ Yes, we do not care about what happens at $(0,1)$, but you need to know that the inequality you used is well defined on $\mathbb R^2 \setminus \{(0,1)\}$. You verified the inequality for the points where $x\neq 0$ and the inequality works just fine when $x=0$ and $y \neq 1$. $\endgroup$ – Hugo C Botós Sep 20 '17 at 4:10

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