Jason Farmer

2021-01-31

Nonexistence of a limit Investiage the limit $\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{\left(x+y\right)}^{2}}{{x}^{2}+{y}^{2}}$

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By applying identity ${\left(a+b\right)}^{2}=\left({a}^{2}+{b}^{2}+2ab\right)$
$\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{\left(x+y\right)}^{2}}{{x}^{2}+{y}^{2}}=\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}+{y}^{2}+2xy}{{x}^{2}+{y}^{2}}$
$\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{\left(x+y\right)}^{2}}{{x}^{2}+{y}^{2}}=\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}+{y}^{2}}{{x}^{2}+{y}^{2}}+\frac{2xy}{{x}^{2}+{y}^{2}}$
$\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{\left(x+y\right)}^{2}}{{x}^{2}+{y}^{2}}=\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}+{y}^{2}}{{x}^{2}+{y}^{2}}+\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{2xy}{{x}^{2}+{y}^{2}}$
$\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{\left(x+y\right)}^{2}}{{x}^{2}+{y}^{2}}=\underset{\left(x,y\right)\to \left(0,0\right)}{lim}1+\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{2xy}{{x}^{2}+{y}^{2}}$
$\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{\left(x+y\right)}^{2}}{{x}^{2}+{y}^{2}}=1+0$
$\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{\left(x+y\right)}^{2}}{{x}^{2}+{y}^{2}}=1$

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