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How to get the following limit:

$$\lim_{(x,y)\to (0,0)}\frac{x^4y}{x^8+y^2}=?$$

If I let $x=r\cos \theta$ and $y=r\sin \theta$ where $\theta\in (0, \pi/2)$, then $$\lim_{(x,y)\to (0,0)}\frac{x^4y}{x^8+y^2}=\frac{r^5\cos^4\theta\sin\theta}{r^8\cos^8\theta+r^2\sin^2\theta}$$

It seems the limit does not exist.

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  • $\begingroup$ Hint : Put $y=tx^4$ $\endgroup$
    – Kelenner
    Aug 17, 2020 at 8:50

2 Answers 2

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In this cases, often a good strategy is to use a change of variable to make the exponents equal at the denominator, indeed let $x^4=u$ and $y=v$ then

$$\lim_{(x,y)\to (0,0)}\frac{x^4y}{x^8+y^2}=\lim_{(u,v)\to (0,0)}\frac{uv}{u^2+v^2}$$

and we can easily conclude for example by polar coordinates or assuming two different paths as $u=\pm v$.

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Along the curve $y=x^{4}$ the limit is $\frac 1 2 $ and along $y=0$ it is $0$. Hence the limit does not exist.

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