# Solving $\lim\limits_{(x,y)\to(0,0)}\;\frac{x^5 + \,y^5}{x^3+\,y^3}$

How do I solve the limit $$\lim_{(x,y)\to(0,0)}\;\frac{x^5+y^5}{x^3+y^3}\quad ?$$ I have tried using polar coordinates, but I don't think an answer would be valid because theta is not fixed. What else can I do?

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You can pass to polar coordinates and, for every theta, make the radius approach zero and see what happens. – Marra Apr 2 '12 at 19:45
Can you explain what you mean? Could you show me an example? – shmiggens Apr 2 '12 at 19:50
Well, instead of trying to calculate this limit, why don't you try to show that it doesn't exist first? When you go to polars, you could try (or another similar coordinate change) $x=r\cos(t)$ and $y=r\sin(t)$. Now you can choose two differents values for t, let's say, $t_1$ and $t_2$ and then calculate the limit when $r\rightarrow 0$ for both values of t. If obtain two different limits, then the limit doesn't exist (why?). The needed cleverness of this is finding values for t that works. – Marra Apr 2 '12 at 19:55
How is that different from what I thought I could not do? I don't think polar can be used at all in this problem. I also don't think that a linearization of y can be used such that y=mx. I don't think an answer would be valid based on either reasoning. – shmiggens Apr 2 '12 at 20:10
Why can't you do it? It seems reasonable enough; a function like that has a limit a if when x reaches b it has a limit over all curves that reaches b. If it doesn't then the limit doesn't exist. – Marra Apr 2 '12 at 20:12

Hint

$$\frac{x^5+y^5}{x^3+y^3} =\frac{x^5+x^2y^3}{x^3+y^3}+ \frac{x^3y^2+y^5}{x^3+y^3} - \frac{x^2y^2(x+y)}{x^3+y^3}$$

And

$$x^2-xy+y^2 \geq |xy| \,.$$

OK, to make it more clear. If you combine the two hints, you get:

$$\left|\frac{x^5+y^5}{x^3+y^3} \right| \leq x^2+y^2+|xy| \,.$$

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I don't understand your hint. – shmiggens Apr 2 '12 at 20:38
@shmiggens: The first fraction should be $\frac{x^5+y^5}{x^3+y^3}$. The three terms on the righthand side of the equation simplify to $$x^2+y^2-\frac{x^2y^2}{x^2-xy+y^2}\;.$$ – Brian M. Scott Apr 2 '12 at 21:10
@BrianM.Scott thank you, missed the $y^5$ there. – N. S. Apr 2 '12 at 21:14
Thank you, to both Brian M. Scott and N.S. – shmiggens Apr 2 '12 at 21:28

Here’s a somewhat brute force argument that doesn’t mention $p$-norms.

Let $u=\dfrac{y}x$, so that $$f(x,y)=\frac{x^5+y^5}{x^3+y^3}=x^2\frac{1+u^5}{1+u^3}\;.$$

Now consider the fraction $\dfrac{1+u^5}{1+u^3}$:

$$\begin{cases} 0<\frac{1+u^5}{1+u^3}\le 1,&\text{ if }0\le u\le 1\;;\\ 1\le\frac{1+u^5}{1+u^3}<\frac53,&\text{ if }-1<u\le 0\;;\text{ and}\\ 0<\frac{1+u^5}{1+u^3}<u^2,&\text{ if }|u|>1\;. \end{cases}$$

(Note that $u$ cannot be $-1$, since $f(x,-x)$ is undefined.) Here the first and third cases are straightforward, and the second is easily checked using l’Hospital’s rule to see what happens as $u\to-1^+$.

It follows that

$$\begin{cases} f(x,y)<\frac53x^2,&\text{if }-1<u\le 1\;,\text{ and}\\\\ f(x,y)<x^2u^2=y^2,&\text{if }|u|>1\;. \end{cases}$$

In all cases, then, $f(x,y)<2(x^2+y^2)$, and clearly $$\lim_{(x,y)\to(0,0)}f(x,y)=0\;.$$

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@anon: Because I’m still asleep. – Brian M. Scott Apr 2 '12 at 20:01
I like your answer, and I may consider this approach; however, I'm wondering if there is a simpler method that is less "brutish" and simpler. – shmiggens Apr 2 '12 at 21:06

Using p-norms, the problem can be rewritten as computing $\lim_{z \rightarrow 0} \frac{||z||_5^5}{||z||_3^3}$. The term in the limit can be written as $||z||_5^2 \frac{||z||_5^3}{||z||_3^3}$, and since $||z||_5 \le ||z||_3$, it follows that $\lim_{z \rightarrow 0} \frac{||z||_5^5}{||z||_3^3} \leq \lim_{z \rightarrow 0} ||z||_5^2 = 0$.

Oops, just realized that my 'proof' assumes $x,y$ are non-negative.

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Thanks for the the effort, at least. – shmiggens Apr 2 '12 at 20:12

The polar co-ordinates approach works except where $x=-y$.

The case where $x=-y$ causes problems as the function is undefined there - not because polar co-ordinates cause the problem. This, however, is removable, since the factor $(x+y)$ can be cancelled from numerator and denominator. Then the denominator becomes:

$$x^2-xy+y^2 = (x-\frac y2)^2 +\frac{3y^2}4$$

which is visibly non-zero except for $x=y=0$. Strictly the original limit is not defined as the function is not defined on the line $y=-x$, and one can approach zero that way. However cancelling $(x+y)$ and setting $y=-x$ gives the function a value of $\frac{5x^2}3$ along this line.

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Does the limit not exist, then? – ahorn Mar 12 '15 at 19:06
@ahorn There is a line where the function in its original form is not defined, so there is no limit at any point along that line. The singularity represented by that line is removeable by cancelling the factor $x+y$ - if that is an allowable operation you can recover a limit which makes sense over the whole plane – Mark Bennet Mar 12 '15 at 22:28