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 $x_2^2<x_1^4$. 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|\\ &<x_1^2. \end{eqnarray} Now it is easy to use the $\epsilon$-$\delta$ definition.

  • $\begingroup$ I think it must be $x_1^4$ instead of $x^4$ in the first line. $\endgroup$
    – Philipp
    May 5 '20 at 18:57
  • $\begingroup$ @Philipp, it is a typo. $\endgroup$
    – xpaul
    May 5 '20 at 19:36
  • $\begingroup$ How did you come up with this idea/algebraic manipulations? Is it just mathematical experience one gains over time or did you know this problem? $\endgroup$
    – Philipp
    May 5 '20 at 19:49
  • $\begingroup$ It is just math experience. I didn't know this problem before. $\endgroup$
    – xpaul
    May 5 '20 at 23:04

Write $f(x)=1/\sqrt {1+g(x)}$ and examine $g(x)=x_2\cdot \frac {x_2}{(x_1)^2}.$


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