Is $|x+y|/\sqrt{x^2+y^2}$ continuous?

Question

Let $g: \mathbb{R}^2 $ -> $\mathbb{R}$

Can I just do the following?

$x^2+y^2\geq 2|x+y|$, so $\sqrt{x^2+y^2}\geq \sqrt{2|x+y|}$, therefore $$\left|\frac{x+y}{\sqrt{x^2+y^2}}\right|\leq\frac{|x+y|}{\sqrt{2|x+y|}}=\frac{\sqrt{|x+y|}}{\sqrt{2}},$$ which converges to $0$ as $x,y\to 0$, that means that it is continuous.

Or is that wrong?

Answer 1

Hint:

Set $y=mx$. For all $x,y\ne0$, the function is

$$\frac{|1+m|}{\sqrt{1+m^2}},$$ which is not constant.

Answer 2

HINT

Note that

$$x^2+y^2\geq 2|x+y|$$

is not true for $x=y=\frac12$.

Check the path with $x=-y=t\to 0$.

Answer 3

You wrote that $$x^2+y^2>2|x+y|$$

which is false. Take $x=y=\frac12$. it gives

$\frac 12>2$.

Answer 4

Do not overthink this. When $x=0, y\ne 0$ the function equals $1$, but if instead $y=-x$ with both of these arguments nonzero then the function equals $0$. There is no single value for $f(0,0)$ that renders the function continuous because the algebraic expression for other values of $x$ and $y$ does not have a (single, invariant) limit as $(x,y)$ approaches $(0,0)$.