I tried to proof a theorem that is wrong.
Now am trying to figure out what's wrong with my proof.
Theorem :- If f(x,y) is a real valued function , such that it has the same limit L at (0,0), along every line y=m*x , then it has limit L at (0,0).

  • for any $\epsilon >0$ , no matter however small, $\exists$ an $\delta >0$ such that $\mid f(x,mx)-L \mid < \epsilon $ whenever $\sqrt{x^2(1+m^2)} < \delta$
  • Above statement holds for any $m$.
  • Choose any $\epsilon$ , say $\epsilon_0$, now corresponding to this :-
  • Let the set of all $\delta$ corresponding to various $m$ be $\{\delta(m)\}$ and let $\xi(\epsilon_0) = $ inf{δ(m)|m∈R} :-
  • for all $m$ , $\mid f(x,mx)-L \mid < \epsilon $ whenever $\sqrt{x^2(1+m^2)} < \xi(\epsilon_0)$

  • Now consider the region $ R=\{(x,y) : \sqrt{x^2+y^2} < \xi(\epsilon_0))\}$

  • Corresponding to any element of this set an $m$ can be found such that $y=mx$
  • which means whole region is same as the region $\sqrt{x^2(1+m^2)} < \xi(\epsilon_0)$

The proof is completed using 5th point and definition of limit.
The case of (x,y) approaching (0,0) along X-axis is considered separately.
Please help me in finding what is wrong in this proof.


For each $\varepsilon>0$, you take a $\delta>0$ such that… But the $\delta$ should actually be written as $\delta(m)$, since it depends upon the choice of $m$. And then you take the minimum of all $\delta(m)$. That's where the problem lies. In general, the minimum doesn't exist and furthermore $\inf\{\delta(m)\,|\,m\in\mathbb{R}\}$ may well be $0$.

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  • $\begingroup$ if minimum does not exist we can take inf{δ(m)|m∈R} as $\xi$ . so the only problem is the case where inf{δ(m)|m∈R} is 0 for some $\epsilon$ ?am I right? $\endgroup$ – Jeevesh Juneja Oct 22 '18 at 12:03
  • 1
    $\begingroup$ @KavitaJuneja Yes, you are right. $\endgroup$ – José Carlos Santos Oct 22 '18 at 12:29

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