Let $M:= \{x = (x_1,x_2) \in \mathbb{R}^2 \; | \; x_1 > \sqrt{|x_2|}\}$ and define
$f: M \to \mathbb{R}$ by $f(x):= \frac{x_1}{\|x\|_2}$ for $x = (x_1,x_2) \in M$, where $\|\cdot\|_2$ means the Euclidean norm.
Show, that f can be extended continuously in $(0,0)$.
So my guess is that $f$ can be extended continuously to the value "1". Therefore, I have to use the $\epsilon-\delta$-criterion and hence I have to show $\|f(x) - 1\| < \epsilon$ for any given $\epsilon > 0$ and assumption $\|x\| < \delta$.
Because $\|x\|_2 \ge x_1$ and $x_1 > \sqrt{|x_2|}$, I get
$|f(x) - 1| = \left|\frac{x_1}{\|x\|_2} - 1\right| = \left|\frac{x_1 - \|x\|_2}{\|x\|_2}\right| = \frac{\|x\|_2 - x_1}{\|x\|_2} < \frac{\|x\|_2 - \sqrt{|x_2|}}{\|x\|_2}$
How can the resulting term be estimated, so that $\frac{\|x\|_2 - \sqrt{|x_2|}}{\|x\|_2} < \epsilon$ under the assumption $\|x\|_2 < \delta$ or maybe $\|x\| < \delta$ for any other norm?
I will write $(x,y)$ instead.
\begin{align*} \left|\dfrac{x}{\sqrt{x^{2}+y^{2}}}-1\right|&=\left|\dfrac{x-\sqrt{x^{2}+y^{2}}}{\sqrt{x^{2}+y^{2}}}\right|\\ &=\left|\dfrac{-y^{2}}{\sqrt{x^{2}+y^{2}}(x+\sqrt{x^{2}+y^{2}})}\right|, \end{align*} note that \begin{align*} \dfrac{|y|}{\sqrt{x^{2}+y^{2}}}\leq 1 \end{align*} for $(x,y)\ne(0,0)$.
Therefore, for $(x,y)\in M$, $x>\sqrt{|y|}\geq 0$ and hence \begin{align*} \left|\dfrac{x}{\sqrt{x^{2}+y^{2}}}-1\right|&\leq\dfrac{|y|}{x+\sqrt{x^{2}+y^{2}}}\\ &\leq\dfrac{|y|}{\sqrt{|y|}+\sqrt{|y|+y^{2}}}\\ &\leq\dfrac{|y|}{\sqrt{|y|}+\sqrt{|y|}}\\ &=\dfrac{1}{2}\sqrt{|y|}\\ &\rightarrow 0 \end{align*} as $(x,y)\rightarrow(0,0)$, $(x,y)\in M$.