# Existence of limit of $f(x)=\frac{x_1}{\Vert x\Vert_2}$ at ${0 \choose 0}$

We are given the function:

$$f: M\subset\mathbb{R}^2 \to \mathbb{R}$$, where $$f(x)=\frac{x_1}{\Vert x\Vert_2}$$ and $$M:=\{x={x_1 \choose x_2}\in\mathbb{R}^2~:~x_1>\sqrt{|x_2|}\}$$.

Show that the limit at $${0 \choose 0}$$ exists.

I already figured out that the limit must be $$1$$.

As the domain is restricted to specific points I could not properly use zero-sequences to show the existence of the limit. So I tried to apply th $$\epsilon$$-$$\delta$$-criterion for limits. However, I could not find an upper boundary for $$\left|\frac{x_1}{\Vert x\Vert_2}-1\right|$$ such that for all $$x \in M$$ with $$\Vert x - 0\Vert <\delta$$ we get: $$\left|\frac{x_1}{\Vert x\Vert_2}-1\right|\leq....<\epsilon$$. At the begining I was optimistic to get such an upper boundary if I incorporate the condition of $$M$$ but it didn't get me anywhere...

Maybe there is some secret trick...

As this is homework I would appreciate if you just provide me a little hint and not the full solution unless you hide it.

Note $$x_1>0$$ and $$x_2^2. So $$\begin{eqnarray} |f(x)-1|&=&\bigg|\frac{x_1}{\sqrt{x_1^2+x_2^2}}-1\bigg|\\ &=&\bigg|\frac{\sqrt{x_1^2+x_2^2}-x_1}{\sqrt{x_1^2+x_2^2}}\bigg|\\ &=&\bigg|\frac{x_2^2}{\sqrt{x_1^2+x_2^2}(\sqrt{x_1^2+x_2^2}+x_1)}\bigg|\\ &<&\bigg|\frac{x_1^4}{\sqrt{x_1^2+x_2^2}(\sqrt{x_1^2+x_2^2}+x_1)}\bigg|\\ & Now it is easy to use the $$\epsilon$$-$$\delta$$ definition.
• I think it must be $x_1^4$ instead of $x^4$ in the first line. May 5 '20 at 18:57
Write $$f(x)=1/\sqrt {1+g(x)}$$ and examine $$g(x)=x_2\cdot \frac {x_2}{(x_1)^2}.$$