proof regarding limit of function of 2 variable along different paths

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).
PROOF :-

• 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.

1 Answer

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 $$0$$.

• 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? – Jeevesh Juneja Oct 22 '18 at 12:03
• @KavitaJuneja Yes, you are right. – José Carlos Santos Oct 22 '18 at 12:29