# Something is wrong with this proof, limits

Could someone please tell me what is wrong with this proof?

Show that $\lim\limits_{(x,y) \to (0,0)} \dfrac{xy^3}{x^4 + 3y^4}$ does not have a limit or show that it does and find the limit.

I know it is wrong because the limit doesn't exist, but this proof is contradicing me

Proof

Case 1

Assume for $x,y > 0$, then $x^4 + 3y^4 > x^4 > x > 0$

\begin{align*} 0 < x < x^4 + 3y^4 &\iff 0 < \dfrac{x}{x^4 + 3y^4} < 1 \\ & \iff 0 < \dfrac{x|y^3|}{x^4 + 3y^4} < |y^3| \\ & \iff \lim\limits_{(x,y) \to (0,0)} 0 < \lim\limits_{(x,y) \to (0,0)} \dfrac{x|y^3|}{x^4 + 3y^4} < \lim\limits_{(x,y) \to (0,0)}|y^3|\\ &\iff 0 < \lim\limits_{(x,y) \to (0,0)} \dfrac{x|y^3|}{x^4 + 3y^4} < 0 \end{align*}

Case 2. WLOG Assume $x,y <0$ and combine both cases.

What's wrong the pproof? I don't find the flaw

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Marvis has mentioned the main errors in his answer, but there’s one more: when you pass to the limits, you have to change $<$ to $\le$. As it stands, your last line says that $0<0$, which is clearly wrong. –  Brian M. Scott May 25 '12 at 5:19
Ahhhhh! okay thanks! –  sidht May 25 '12 at 5:26
When $x=y\ne 0$, the function is $1/4$, so estimates cannot be right. When $x=-y\ne 0$ the function is $-1/4$, so limit cannot exist. –  André Nicolas May 25 '12 at 5:31

The statement $x^4>x$ is clearly wrong especially when $x$ is in the neighborhood of $0$.

Also, you need to look at cases $x>0,y<0$ and $x<0,y>0$. But the main error is the statement $x^4 > x$.

EDIT

The answer is that the function is not continuous at the origin. This can be seen as follows. Remember that in two dimensions there are infinitely many different directions from which you can approach origin unlike in one dimension where you need to look only at two different cases $x>0$ and $x<0$. Hence, in two dimensions it is not sufficient/advantageous to split it into just four cases $x,y>0$, $x,y<0$, $x>0,y<0$, $x<0,y>0$ and analyze.

Approach the origin along the straight line $y=mx$. Then we have that $$\lim_{\overset{x \rightarrow 0}{y \rightarrow 0}} \dfrac{xy^3}{x^4 + 3y^4} = \lim_{\overset{x \rightarrow 0}{y = mx}} \dfrac{xy^3}{x^4 + 3y^4} = \lim_{x \rightarrow 0} \dfrac{x(mx)^3}{x^4 + 3(mx)^4} \\ = \lim_{x \rightarrow 0} \dfrac{m^3 x^4}{x^4 + 3m^4 x^4} = \lim_{x \rightarrow 0} \dfrac{m^3 x^4}{(1 + 3m^4) x^4} = \lim_{x \rightarrow 0} \dfrac{m^3}{(1 + 3m^4)} = \dfrac{m^3}{1+3m^4}$$ Hence, approaching the origin along different straight lines give different answers.

For instance, if we approach the origin along the line $y = x$, the limit is $\dfrac14$.

If we approach the origin along the line $y = -x$, the limit is $-\dfrac14$.

If we approach the origin along the line $y = 0$, the limit is $0$.

This proves that the function is not continuous at the origin.

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I realize the proof is wrong, but I want to understand the technique. I know it's kinda hard to do with this function since it has no limit anyways, but if I were to go with x,y>0, how do continue with my proof? –  sidht May 25 '12 at 5:30
@jak I have updated my answer and hope it clarifies your question. –  user17762 May 25 '12 at 5:38
@jak: The $x$, $y$ positive part doesn't really help, approach $(0,0)$ along $y=x$, or $y=2x$, get different answers. –  André Nicolas May 25 '12 at 5:39
@jak Remember that in two dimensions there are infinitely many different directions from which you can approach origin unlike in one dimension where you need to look only at two different cases $x>0$ and $x<0$. Hence, in two dimensions it is not sufficient/advantageous to split it into just four cases $x,y>0$, $x,y<0$, $x>0,y<0$, $x<0,y>0$ and analyze. –  user17762 May 25 '12 at 5:44
One cannot simply look at approaching the origin along a line. Approaching the origin along any line, $$\frac{x^2-y}{(x^2+y^2)^2+x^2-y}$$ limits to $1$. However, approaching along the curve $y=x^2$, the limit is $0$. –  robjohn May 25 '12 at 9:28
show 1 more comment

The easiest way to see that a limit doesn't exist is to write $$\frac{xy^3}{x^4+y^4}=\frac{x/y}{(x/y)^4+1}\tag{1}$$ Using $(1)$, it is easy to see that, over any circle around the origin, $\frac{xy^3}{x^4+y^4}$ can take on any value that $\frac{t}{t^4+1}$ can; that is, any value in $\left[{-}\sqrt[4]{\frac{27}{256}}\;,\;\sqrt[4]{\frac{27}{256}}\right]$. Therefore, $$\lim_{x,y\to0}\frac{xy^3}{x^4+y^4}\tag{2}$$ does not exist.

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In the first line of your proof, $x^4$ is not greater than $x$ just because $x$ is positive. For numbers smaller than $1$, this is false.

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Try a different approach. Literally.

If the limit exists, then it will be the same regardless of how you approach $(0,0)$.

So, consider the limit first along the line $y=x$, and you will see that as long as $x\neq 0$, then $\frac{xy^3}{x^4 + 3y^4} = \frac{1}{4}$. Now try approaching along the line $y=2x$, and see what limit you get. If they are different, then this is proof that the limit as $(x,y) \to (0,0)$ doesn't exist.

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Just adding to the answers. When you have limits with several variables it is very useful to try to visualize it first.
When you see that different values are trying to "get closer and closer" to the same point, then you might expect that something fishy is going on.

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