I know the limit of the following functions at $(0,0)$ are both $0$. But how do I prove them?
a) $f(x,y)=\frac{x^3-2y^3}{x^2+2y^2}$
b) $f(x,y)=\frac{xy^4}{x^2+y^6}$
Limit of multivariate function at $(0,0)$.
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limits
1 Answers
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For the first approach with ellipses, $x=r\cos (\theta)$ and $y=\frac{1}{\sqrt{2}}r\sin (\theta)$ with $r \to 0^+$, regardless of $\theta$.
$$=\lim_{r \to 0^+} \frac{r^3 \cos^3 (\theta)-\frac{1}{\sqrt{2}}r^3 \sin^3 (\theta)}{r^2}$$
For the second limit, looking at denominator set it equal to, $r^2$. So $x^2+y^6=r^2$ which gives some wierd but closed shape. Now use $x=r \cos (\theta)$ and $y=r^{1/3} \sin^{1/3} (\theta)$ . And approach $r \to 0^+$ regardless of $\theta$.
$$=\lim_{r \to 0^+} \frac{r^{7/3} \cos (\theta) \sin^{4/3} (\theta)}{r^2}$$