Path dependcy and delta epsilon proofs

So I have this major confusion with multi-variable delta epsilon proofs.

Basically, I have this example:

lim(x,y)→(0,0)x4−4y2x2+2y2\lim_{(x,y) \rightarrow (0,0)} \frac{x^4 – 4y^2}{x^2 + 2y^2}

Now by path dependency, we have the limit doesn’t exist:

Suppose y=mxy = mx.

Then: x4−4y2x2+2y2=x4−4m2x2x2+2m2x2=x2−4m21+2m2 \frac{x^4 – 4y^2}{x^2 + 2y^2} = \frac{x^4 – 4m^2x^2}{x^2 + 2m^2x^2} = \frac{x^2 – 4m^2}{1 + 2m^2}

Thus as x→0x \rightarrow 0, the limit is path reliant and doesn’t exist.

However if I try to do a delta epsilon proof:

Fix ϵ>0\epsilon > 0 and suppose x =(x,y)= (x, y).

Suppose 0<||0 < ||x||<δ|| < \delta. Note x4−4y2x2+2y2≤||x||4−4||x||2||x||2+2||x||2≤||x||2−41+2<||x||23<δ23\frac{x^4 - 4y^2}{x^2 + 2y^2} \leq \frac{||\textbf{x}||^4 - 4||\textbf{x}||^2}{||\textbf{x}||^2 + 2||\textbf{x}||^2} \leq \frac{||\textbf{x}||^2 - 4}{1 + 2} < \frac{||\textbf{x}||^2}{3} < \frac{\delta^2}{3} Let δ=3√ϵ\delta = 3\sqrt\epsilon. What's wrong with the above? Doesn't that imply the limit equals 0? =================      Your fourth line from below, is incorrect. and even, Why not δ2\delta^2? – Abdallah Hammam 2 days ago 2   You incorrectly used your triangle inequality : ||x4−4y2||≤||x4||+4||y2||||x^4-4y^2||\le ||x^4||\mathbf{+}4||y^2||. – Nicolas FRANCOIS 2 days ago      |x|≠|y||x|\ne |y| in general. And you've misapplied the triangle inequality. – Dr. MV yesterday ================= =================