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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 don't *solve* a limit, you *evaluate* it.2012-04-02

4 Answers 4

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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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    @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 plane2015-03-12
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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 $\dfrac{1+u^5}{1+u^3}$:

$\begin{cases} 0<\frac{1+u^5}{1+u^3}\le 1,&\text{ if }0\le u\le 1\;;\\ 1\le\frac{1+u^5}{1+u^3}<\frac53,&\text{ if }-11\;. \end{cases}$

(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

$\begin{cases} f(x,y)<\frac53x^2,&\text{if }-11\;. \end{cases}$

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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    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.2012-04-02
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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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    Thank you, to both Brian M. Scott and N.S.2012-04-02
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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.2012-04-02