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$$\lim_{(x,y)\to(0,0)}\frac{ x ^ 2 + 3 x - 4 y }{ x - y}$$

My try:

Since the limit does not exist when we substitute 0,0 I thought of proving that limit does not exists.

$y=0,x \to 0^{+}$

$f(x,0)=x+3$

$x=0,y \to 0^+$

$f(0,y)=4$

Since $f(x,0)\neq f(0,y)$ then limit does not exist.

Is my attempt correct?

Thanks..

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  • $\begingroup$ You can obtain $\to$ by typing $to$. $\endgroup$ – N. F. Taussig Sep 27 '18 at 10:28
  • $\begingroup$ Though not written with perfect rigor, you are right. $\endgroup$ – Yves Daoust Sep 27 '18 at 10:35
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You mean $$\lim_{x \to 0} f(x,0) = 3 \ne 4 = \lim_{y \to 0} f(0,y).$$

Since they are not equal when we travel along two trajectories, the limit doesn't exist.

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With $m:=\dfrac yx$, you have

$$\frac{ x ^ 2 + 3 x - 4 y }{ x - y}=\frac{ x + 3 - 4 m }{ 1 - m}\to\frac{3-4m}{1-m}$$

which is not a constant function of $m$.

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