 # How can I prove that $$\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}{\frac{{{\left|{x}\right|}^{{{\frac{{{3}}}{{{2}}}}}}{y}^{{{2}}}}}{{{x}^{{{4}}}+{y}^{{{2}}}}}}\rightarrow{0}$$ Lesnaoq73 2022-03-27 Answered
How can I prove that
$\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{|x|}^{\frac{3}{2}}{y}^{2}}{{x}^{4}+{y}^{2}}\to 0$
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I assume you want to show that
$\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{|x|}^{\frac{3}{2}}{y}^{2}}{{x}^{4}+{y}^{2}}\to 0$
Although sometimes it is not the nicest way, switching to polar coordinates is a general approach that will solve these kinds of limits. Let . Then your limit is
$\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{|x|}^{\frac{3}{2}}{y}^{2}}{{x}^{4}+{y}^{2}}=\underset{r\to 0}{lim}\frac{{r}^{\frac{3}{2}}{|\mathrm{cos}\theta |}^{\frac{3}{2}}{r}^{2}{\mathrm{sin}}^{2}\theta }{{r}^{2}\left({r}^{2}{\mathrm{cos}}^{4}\theta +{\mathrm{sin}}^{2}\theta \right)}=\underset{r\to 0}{lim}{r}^{\frac{3}{2}}\frac{{|\mathrm{cos}\theta |}^{\frac{3}{2}}}{\left({r}^{2}\frac{{\mathrm{cos}}^{4}\theta }{{\mathrm{sin}}^{2}\theta }+1\right)}$
Now, since
$0\le \frac{{|\mathrm{cos}\theta |}^{\frac{3}{2}}}{\left({r}^{2}\frac{{\mathrm{cos}}^{4}\theta }{{\mathrm{sin}}^{2}\theta }+1\right)}\le {|\mathrm{cos}\theta |}^{\frac{3}{2}}\le 1$
we can apply the squeeze theorem, and since $\underset{r\to 0}{lim}{r}^{\frac{3}{2}}=0$, we see that the original limit is 0.
Hope that helps.