# Prove whether the limit exists where $(x,y)\to(0,0)$, $f(x,y)=\frac{x^4+y^4}{x^3+y^3}$.

Prove whether the limit exists where $(x,y)\to(0,0)$, $f(x,y)=\dfrac{x^4+y^4}{x^3+y^3}$.

After doing polar coordination i get the next expression $\displaystyle\lim_{r\to 0}r\frac{\cos^4x+\sin^4x}{\sin^3x+\cos^3x}$ I thought this one should be equal to $0$.

But i seems that the limitation above does no exists, How do i prove that one then?

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You can choose to approach $(0, 0)$ along the line $x = -y$ to see that the limit does not exist on that line. This corresponds to the case when $\sin^3\theta + \cos^3\theta = 0$. – Tunococ Aug 1 '13 at 12:36
Why not just put $y=0$ and take the limit of $x$ and vice-versa? – Graham Hesketh Aug 1 '13 at 12:39
@GrahamHesketh, because that's just one way to evaluate the limit (i.e., along the line $\,y=0\,$) . If the limit exists it must exist no matter how we approach the origin... – DonAntonio Aug 1 '13 at 12:50

What happens in your limit if

$$\sin x=-\cos x\iff \tan x=-1\iff x=-\frac\pi4+k\pi\;,\;\;k\in\Bbb Z\;\;?$$

This gives us the idea to put $\,y=-x \;$ and see what happens when $\,x\to 0\;$:

$$\lim_{x\to 0}\frac{x^4+y^4}{x^3+y^3}=\lim_{x\to 0}\frac{2x^4}{x^3-x^3}$$

Since the function isn't even defined on $\,y=-x\,$ the limit cannot exist (as it must exist no matter how we approach the origin!)

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...but the numerator is also $0$ at the origin? So the limit may be finite and exist at that point as in e.g. the sinc function. – Graham Hesketh Aug 1 '13 at 12:47
We don't care what happens to some function at the point where the variable(s) are going...in fact, by definition, we must not care. The point here is that the function isn't defined at all in the whole line $\,y=-x\,$ and thus cannot approach the origin along this line and thus the wanted limit, by definition, cannot exist. – DonAntonio Aug 1 '13 at 12:49
Ok fair enough... :) – Graham Hesketh Aug 1 '13 at 12:50
(+1) Just found this: sinclair.edu/centers/mathlab/pub/findyourcourse/worksheets/…, it's a shame though because the limit is the same from any other line of approach in the whole domain!! :( lolz – Graham Hesketh Aug 1 '13 at 12:54
Indeed so @GrahamHesketh – DonAntonio Aug 1 '13 at 16:33

more easy way is $y=ax$,then $f(x,y)=\dfrac{x^4+y^4}{x^3+y^3}=x\dfrac{1+a^4}{1+a^3}$, so the answer is $0$

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...and what happens if $\,a=-1\,$ ?! – DonAntonio Aug 1 '13 at 12:45
but it is not allowed $a=-1$ according to $f(x.y)$ – chenbai Aug 7 '13 at 6:01
I don't understand what you say, @chenbai: the definition of $\,f\,$ doesn't mention any $\,a\,$, it is you who defined it, and by definition of the limit it must exist no matter how $\,(x,y)\to(0,0)\,$ , and one way it is $\,y=-x\,$ , and the function isn't defined here, so the limit cannot exist. – DonAntonio Aug 7 '13 at 6:14
what I was taught this function can't allow y=-x in anyway. so I think I can exclude this case. If I can't do it, I do learn a new idea. pls confirm me.( I saw your answer). thanks. – chenbai Aug 7 '13 at 6:35
thanks. now I understand when I calculate the limit I have to care all cases. I am happy to know that and thanks a lot. – chenbai Aug 7 '13 at 6:48