Understanding the Limit of 2 Variables (x,y) -> (0,0) for (x^2)(y)/((x^4)+(y^2))

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What is:
lim(x,y)->(0,0) of (x^2)(y) / ((x^4) + (y^2)) ?

When I take x_n = 0, y_n = 1/n, lim=0
and x_n = 1/n, y_n = 0, lim=0
and x_n = y_n = 1/n, lim=0
All three limits are zero, yet other people I've asked say the limit doesn't
exist. Am I right, or am I doing something wrong here? Thanks.
 
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\lim_{(x,y) \rightarrow (0,0)}\frac{x^2y}{x^4+y^2}x=0
\lim_{(x,y) \rightarrow (0,0)}\frac{0\cdot y}{0+y^2}=0y=0
\lim_{(x,y) \rightarrow (0,0)}\frac{x^2\cdot 0}{x^4+0}=0y=x^2
\lim_{(x,x^2) \rightarrow (0,0)}\frac{x^4}{2x^4}=\frac 1 2

Aim to make your powers the same, use y=x^2 or x=y^2.

General tests:

x=y=0
y=x
x=y^n
y=x^n
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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