# Determine the limits if they exist: lim_{(x,y)rightarrow(2,4)}frac{(x-2)^2(y-4)^2}{(x-2)^3+(y-4)^3}

Determine the limits if they exist: $\underset{\left(x,y\right)\to \left(2,4\right)}{lim}\frac{\left(x-2{\right)}^{2}\left(y-4{\right)}^{2}}{\left(x-2{\right)}^{3}+\left(y-4{\right)}^{3}}$
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Given that $\underset{\left(x,y\right)\to \left(2,4\right)}{lim}\frac{\left(x-2{\right)}^{2}\left(y-4{\right)}^{2}}{\left(x-2{\right)}^{3}+\left(y-4{\right)}^{3}}$
Take
$\left(x-2\right)=X$ and $4=Y$ then
when $x\to 2$ then $X\to 0$
when $y\to 4$, then $Y\to 0$
So, the equation (1) becomes
$\underset{\left(x,y\right)->\left(2,4\right)}{lim}\frac{\left(x-2{\right)}^{2}\left(y-4{\right)}^{2}}{\left(x-2{\right)}^{3}+\left(y-4{\right)}^{3}}=\underset{\left(X,Y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}\cdot {y}^{2}}{{X}^{3}\cdot {Y}^{3}}$
To check is limit path dependent or not, so put Y=mX
$⇒\underset{\left(X,Y\right)->\left(0,0\right)}{lim}\frac{{X}^{2}\cdot {Y}^{2}}{{X}^{3}+{Y}^{3}}=\underset{X\to 0}{lim}\frac{{X}^{2}\cdot {m}^{2}\cdot {X}^{2}}{{X}^{3}+\left(mX{\right)}^{3}}=\underset{X->0}{lim}\frac{{m}^{2}\cdot {X}^{4}}{{X}^{3}\left(1+{m}^{3}\right)}$
$⇒\underset{\left(X,Y\right)->\left(0,0\right)}{lim}\frac{{X}^{2}\cdot {Y}^{2}}{{X}^{3}+{Y}^{3}}=\underset{X\to 0}{lim}\frac{{m}^{2}\cdot {X}^{4}}{{X}^{3}\left(1+{m}^{3}\right)}=0$
$\underset{\left(x,y\right)->\left(2,4\right)}{lim}\frac{\left(x-2{\right)}^{2}\left(y-4{\right)}^{2}}{\left(x-2{\right)}^{3}+\left(y-4{\right)}^{3}}=\underset{\left(X,Y\right)->\left(0,0\right)}{lim}\frac{{x}^{2}\cdot {y}^{2}}{{X}^{3}\cdot {Y}^{3}}=0$
Hence, limit exist and
$\underset{\left(x,y\right)\to \left(2,4\right)}{lim}\frac{\left(x-2{\right)}^{2}\left(y-4{\right)}^{2}}{\left(x-2{\right)}^{3}+\left(y-4{\right)}^{3}}=0$