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?
 A: 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 $$\lim_{(x,y)\to(0,0)}f(x,y)=0\;.$$
A: 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:
$$\left|\frac{x^5+y^5}{x^3+y^3}  \right| \leq x^2+y^2+|xy| \,.$$
A: 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.
A: 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.
