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Question Number 68354 by TawaTawa last updated on 09/Sep/19

$$\underset{x,\mathrm{y}\to (0,0)}{\lim }\frac{{\mathrm{x}}^4-{\mathrm{x}}^2{\mathrm{y}}^2+{\mathrm{y}}^4}{{\mathrm{x}}^2+{\mathrm{x}}^4{\mathrm{y}}^4+{\mathrm{y}}^2}$$

Commented by kaivan.ahmadi last updated on 09/Sep/19

$$y=x\Rightarrow$$$${lim}_{x\to 0}\frac{x^4}{2x^2+x^8}={lim}_{x\to 0}\frac{x^2}{2+x^6}=0$$$$x=0\Rightarrow$$$${lim}_{y\to 0}\frac{y^4}{y^2}=0$$$$y=0\Rightarrow$$$${lim}_{x\to 0}\frac{x^4}{x^2}=0$$$$y=x^n\Rightarrow itiszero$$$$probiblyitiszero.$$$$$$$$$$

Commented by MJS last updated on 09/Sep/19

$$\mathrm{just}\mathrm{ideas}:$$$$\left(1\right)$$$$\underset{t\to 0}{\lim }\frac{t^4-{at}^2+b}{{ct}^4+t^2+d}=\frac{b}{d}$$$$\mathrm{inserting}\mathrm{we}\mathrm{get}\underset{x\to 0}{\lim }(...)=y^2\mathrm{and}\underset{y\to 0}{\lim }(...)=x^2$$$$\Rightarrow \mathrm{if}\mathrm{both}x\to 0\mathrm{and}y\to 0\Rightarrow \mathrm{answer}\mathrm{is}0$$$$$$$$\left(2\right)$$$$\mathrm{let}y=\alpha x\mathrm{with}\alpha \in \mathbb{R}$$$$\frac{x^4-x^2{y}^2+y^4}{x^2+x^4{y}^4+y^2}=\frac{({\alpha }^4-{\alpha }^2+1)x^2}{{\alpha }^4{x}^6+{\alpha }^2+1}$$$$\underset{x\to 0}{\lim }\frac{({\alpha }^4-{\alpha }^2+1)x^2}{{\alpha }^4{x}^6+{\alpha }^2+1}=0$$

Commented by TawaTawa last updated on 09/Sep/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sirs}$$

Commented by mind is power last updated on 09/Sep/19

$$x=rcos\left(a\right)$$$$y=rsin\left(a\right)$$$$x^4-x^2{y}^2+y^4=r^4({cos}^4\left(a\right)+{sin}^4\left(a\right)-{sin}^2\left(a\right){cos}^2\left(a\right))$$$$x^2+y^2+{\left(xy\right)}^4=r^2+r^8{cos}^4\left(a\right){sin}^4\left(a\right)$$$$\frac{x^4+y^4-x^2{y}^2}{x^2+y^2+{\left(xy\right)}^4}=\frac{r^2({cos}^4\left(a\right)+{sin}^4\left(a\right)-{sin}^2\left(a\right){cos}^2\left(a\right))}{1+r^6{\left(cos\left(a\right)sin\left(a\right)\right)}^4}\to 0$$

Commented by TawaTawa last updated on 10/Sep/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by mind is power last updated on 10/Sep/19

$$y,rewelcom$$

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