# Thread: An interesting limit problem

1. ## An interesting limit problem

This should've been an easy problem for me, though a few small details make it tougher than I'd initially thought.

Let $f(x,y)$ be defined as follows: (this is the main part that has me stumped right now)
$f(x,y)=0$ for all $(x,y)$ unless $x^4,
$f(x,y)=1$ for all $(x,y)$ where $x^4.

Show that $f(x,y)\rightarrow 0$ as $(x,y)\rightarrow (0,0)$ on any straight line through $(0,0)$.

Determine if $\lim f(x,y)$ exist as $(x,y)\rightarrow (0,0)$.

If I were to get past that little hurdle in terms of the definition of $f(x,y)$, I could probably get through the rest using epsilon-delta proofs.

2. a) if $(x,y) \rightarrow 0$ along a straight line, then is $y=k\ x$. But if $k > 0$ and $x>0$ [ $k<0$ and $x<0$ leads to similar conclusions...] is $\lim_{x \rightarrow 0} \frac{x^{2}}{k\ x} = 0$ and in 'proximity' of $x=0$ is $k\ x> x^{2}$ so that the limit is 0. If $k =0$ and $x \ne 0$ is $x^{4} > y$ and the limit is again 0...

b) as seen in a) there is almost one 'trajectory' for which the limit is 0. Now we consider the 'trajectory' defined by $y= |x^{3}|$ we find that the limit is 1, so that no limit exists...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by chisigma
a) if $(x,y) \rightarrow 0$ along a straight line, then is $y=k\ x$. But if $k > 0$ and $x>0$ [ $k<0$ and $x<0$ leads to similar conclusions...] is $\lim_{x \rightarrow 0} \frac{x^{2}}{k\ x} = 0$ and in 'proximity' of $x=0$ is $k\ x> x^{2}$ so that the limit is 0. If $k =0$ and $x \ne 0$ is $x^{4} > y$ and the limit is again 0...

b) as seen in a) there is almost one 'trajectory' for which the limit is 0. Now we consider the 'trajectory' defined by $y= |x^{3}|$ we find that the limit is 1, so that no limit exists...

Kind regards

$\chi$ $\sigma$
Thanks for the help. I see how this works out, although I'm probably not going to do so well in showing it properly (well, neatly, at least). And in mathematics, showing your steps is more important than the answer. Bad luck for me, I guess.