3
$\begingroup$

My question is :

Suppose we know that $\lim_{(x,y) \to (a,b)}f(x,y)$ exists in infinitely many paths for a function $f : \mathbb{R}^2\rightarrow \mathbb{R}$ then can we say that limit exists and it can be obtained from choosing any path.

May be i should have been more careful when saying so...

Actual question is to evaluate $\lim_{(x,y) \to (0,0)} \dfrac{x^3y}{x^4+y^2}$

First thing one would do is to check for "Non existence" by which i mean trying out some paths hoping that limit in those two paths would differ and then conclude that limit does not exist.

But then what if the limit in more than one path coincide..

In this situation i would take $y=mx$ then :
$$\dfrac{x^3y}{x^4+y^2}=\dfrac{mx^4}{x^4+m^2x^2}\rightarrow 0 \text { as x $\rightarrow$ 0}$$

We realize that irrespective of $m$ we would end up with $\lim_{(x,y) \to (0,0)} \dfrac{x^3y}{x^4+y^2}=0$ along the path $y=mx$.

I am sure with this we can not say that limit exists...

so, how many paths (I have checked uncountably many as my $m\in \mathbb{R}$) should i check to confirm the existence.

There might be some other ways to calculate the limit by some squeeze principle or some but then Is there any thing that if you verify that limit coincide in this many paths then you can conclude limit exist.

P.S : My question is not about evaluating the limit but actually Existence of limit knowing its value in some paths... so please do not answer with keeping in mind the word "Evaluate"

Thank you :)

$\endgroup$
  • $\begingroup$ IIRC, a possible trick was to use polar coordinates. $\endgroup$ – jinawee Jan 14 '14 at 13:18
  • $\begingroup$ @jinawee : I am not sure if that is sufficient :O $\endgroup$ – user87543 Jan 14 '14 at 13:20
  • 4
    $\begingroup$ If I remember correctly, there is a post on MO showing that if the limit does not exist, then there is a (nice) path along which the limit does not exist. So checking any number of paths, that aren't all of them, doesn't suffice in general. More concretely, Example 9.3 in Gelbaum and Olmsted's Counterexamples in Analysis furnishes an example where the limit does not exist, but the limit along any path of the form $y=cx^m$, where $m$ and $n$ are relatively prime positive integers and $c\ne0$, does. $\endgroup$ – David Mitra Jan 14 '14 at 13:25
  • $\begingroup$ @DavidMitra : could you please cite that MO link if you come across that in a while $\endgroup$ – user87543 Jan 14 '14 at 13:35
  • $\begingroup$ Isn't your question a duplicate of: math.stackexchange.com/questions/307211/… ? $\endgroup$ – jinawee Jan 14 '14 at 13:36
2
$\begingroup$

Here are two examples (Examples 9.1 and 9.2) taken from Gelbaum and Olmsted's Counterexamples in Analysis that illustrate how badly "path checking" can fail:

1) $$ f(x,y)=\cases{ {x^2y\over x^4+y^2 },& if $(x,y)\ne(0,0)$\cr 0,&$(x,y)=(0,0)$} $$

Here the limit of $f$ as $(x,y)$ makes any straight line approach to the origin is $0$. Yet $f$ does not have a limit at $(0,0)$, as there are points arbitrarily near the origin at which $f$ takes the value $1/2$ (namely $(a,a^2)$.

2) $$ f(x,y)=\cases{ {e^{-1/x^2}y\over e^{-2/x^2}+y^2 },& if $x\ne 0$\cr 0,&$x=0$} $$

Here the limit of $f$ as $(x,y)$ makes any approach to the origin along a curve of the form $y=cx^{m/n}$, where $c\ne0$ and $m,n$ are relatively prime positive integers, is $0$. Yet $f$ does not have a limit at $(0,0)$, as there are points arbitrarily near the origin at which $f$ takes the value $1/2$ (namely $(a,e^{-1/a^2})$.

$\endgroup$
  • $\begingroup$ first function is close to being natural... but second one is very far from being natural.. :) Thank you any ways :) $\endgroup$ – user87543 Jan 14 '14 at 14:02

Your Answer

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