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Does this limit exist?

$$\lim_{(x,y)\rightarrow (0,0)}\frac{xy}{2x^2-y^2}$$

Also is there a general method that may be used to answer such questions?

Thanks.

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    Try polar coordinates.2012-11-07
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    sorry I'm unsure what polar cordinates are can you be more specific.2012-11-07
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    well I ran a search and it turns out: I have to let y=rsin(a), and x=rcos(a)2012-11-07
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    Yup, that's right. It's not the *only* way, but it's *a* way.2012-11-07
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    No, $r\to0$. Notice that $r=\sqrt{x^2+y^2}$.2012-11-07
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    Since both numerator and denominator are of same degree, there is no clear dominance and thus they will tightly compete. As a result, the limit achieves different values or even fluctuates according to the avenue of approach and the limit diverges.2012-11-07
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    yup ok thanks :)2012-11-07
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    does $$\lim_{x \to \left(0,0\right)} xy $$ mean as x and y both approach 0?2012-11-07
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    @MaoYiyi yes ...2012-11-07

4 Answers 4

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Along $y=x$ then $xy/(2x^2-y^2)=x^2/2x^2$ has limit 1/2. Along the line $y=2x$ then $xy/(2x^2-y^2)=2x^2/(-2x^2)$ has limit -1. Thus the limit does not exist.

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No it doesn't. Take the test sequence $(0,1/n)$ and the limit is $0$, taking a different test sequence $(1/n,1/n)$ gives the limit $1$.

The general method is to test different, non-trivial directions, usually a sequence along $x=0$ or $y=0$ then a sequence along $y = \pm x$ suffices.

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Choose $\,y=mx\,$ , so that $\,(x,y)=(x,mx)\,$ , and then

$$\frac{xy}{2x^2-y^2}=\frac{mx^2}{x^2(2-m^2)}=\frac{m}{2-m^2}\xrightarrow [(x,y)\to (0,0)\Longleftrightarrow x\to 0]{} \frac{m}{2-m^2}$$

and the limit is not independent of the way the variable approaches the origin, thus it doesn't exist.

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No it does not exist. I will the function in the limit be f(x,y), we find that the sequence: f(0,1/1),f(0,1/2),f(0,1/3),... does not converge, thus we conclude that there is no L such that for all epsilon>0 there exists an open ball B around (0,0) such that for all (x,y) in B |f(x,y)-L|

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    I didnt see the previous answers2012-11-07