zdebe5l8

2022-04-06

I'm trying to solve the following limit:
$\underset{\left(x,y\right)\to \left(2,1\right)}{lim}\frac{xy{\left(x-2\right)}^{\frac{5}{3}}{\left({y}^{2}-1\right)}^{\frac{2}{3}}}{|y|{\left(x-2\right)}^{2}+{x}^{2}{\left(y-1\right)}^{2}}$

szkliwiakinrd

Try writing
$\underset{\left(x,y\right)\to \left(2,1\right)}{lim}\frac{{\left(x-2\right)}^{\frac{5}{3}}{\left({y}^{2}-1\right)}^{\frac{2}{3}}}{{\left(x-2\right)}^{2}+{\left(y-1\right)}^{2}}$
with the substitution $x↦\left(u+2\right)$ and $y↦\left(v+1\right)$ as
$\underset{\left(u,v\right)\to \left(0,0\right)}{lim}\frac{{u}^{\frac{5}{3}}{v}^{\frac{2}{3}}{\left(v+2\right)}^{\frac{2}{3}}}{{u}^{2}+{v}^{2}}$
we know that
${u}^{2}\le {u}^{2}+{v}^{2}$ (1)
and that
$2uv\le {u}^{2}+{v}^{2}$
Multiply (1) to the 1/2 power times (2) to the 2/3 power to get
${2}^{\frac{2}{3}}{u}^{\frac{5}{3}}{v}^{\frac{2}{3}}\le {\left({u}^{2}+{v}^{2}\right)}^{\frac{7}{6}}$
So the whole fraction is $\le {\left({u}^{2}+{v}^{2}\right)}^{\frac{1}{6}}$ and that $\to 0$

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