2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.