# Show that $g(x,y)=\frac{x^2+y^2}{x+y}$ is continuous at $(0,0)$.

Let $g:\mathbb{R^2}\rightarrow \mathbb{R}$ so that, in $M=[0,1]\times[0,1]$,

$$g(x,y)=\begin{cases}\frac{x^2+y^2}{x+y} &\text{ if }x+y \neq 0,\\\\ 0&\text{ if }x+y=0\end{cases}$$

Show that $g$ is continuous at $(0,0)$ in $M$.

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Hint: To find the limit at $(0,0)$ use polar coordinates $x=r\cos(\theta),$ $y=r\sin(\theta)$ and consider taking the limit as $r\to 0$.

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Use polar coordinates. Since $\sin \theta + \cos \theta = \sqrt{2} \sin(\theta + \frac{\pi}{4})$ then for $0 \leq \theta \leq \frac{\pi}{2}$ the denominator is clearly bounded away from zero on $M$.

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In case you do not want to use polar coordinates...

The map $\Phi \colon \mathbb{R}^2 \to \mathbb{R}^2, \ (x,y) \mapsto (x-y,y)$ is a homeomorphism. Let $M' = \Phi^{-1}(M)$, then showing $g$ is continuous on $M$ is equivalent to showing that $g'=g \circ \Phi$ is continuous on $M'$.

Note that for $(a,b) \in M'$ we have that $a \ge b \ge 0$.

Now $$g' (a,b) = \frac{(a-b)^2+b^2}{a-b+b} = a- 2b + \frac{b^2}{a}$$ and using that $a \ge b$ we have the inequality $$g' (a,b)\le a-2b + \frac{b^2}{b}.$$

From this we can see that $g'$ is continuous at $(0,0)$, hence also $g$ is continuos at $(0,0)$.

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