Find values for which the limit exists

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Hi,
I'm given this problem:

Find conditions for variables a, b, c so that the limit

<br /> \lim_{[x,y] \rightarrow [0,0]} \frac{xy}{ax^2 + bxy + cy^2}<br />

exists.

What I have only found so far is that for all variables non-zero the limit doesn't exist. Anyway, I have no clue how to find the conditions for which it does. I tried a = b = c = 0, but it doesn't seem to help to me...

Thank you for the enlightenment.
 
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OOps I was too late in deleting my post. Actually I made a mistake in solving that and yes, what makes me unsure is that I don't think we could use hopital rule for these kind of limit.
 
You'd better to ask another homework helper, but I solve it in another way now . If you look at numerator nad denominator, you see both of them have xy. So?
 
Lisa! said:
You'd better to ask another homework helper, but I solve it in another way now . If you look at numerator nad denominator, you see both of them have xy. So?

Ok, now it seems to me that the condition for the limit to exist is that a = c = 0.

Anyway, it is just the result of guessing method, is there any more exact approach to solve this?
 
For functions like this, where you have two variables, I find it best to convert to polar coordinates. That way, exactly one variable, r, measures the distance to (0,0) which is the crucial factor. In polar coordinates,
x= r cos(\theta) and y= r sin(\theta) so
xy= r^2 cos(\theta)sin(\theta), x<sup>2</sup>= r^2 cos^2(\theta)[/itex], and y^2= r^2 sin^2(\theta)[/itex].&lt;br /&gt; Of course, then ax^2+ bxy+ cy^2= ar^2cos^2(\theta)+ br^2sin(&amp;lt;br /&amp;gt; theta)cos(\theta)+ cr^2sin^2(\theta) so that&lt;br /&gt; ax^2+ bxy+ cy^2= r^2(acos^2(\theta)+ bsin(&amp;lt;br /&amp;gt; theta)cos(\theta)+ csin^2(\theta).&lt;br /&gt; &lt;br /&gt; That means that&lt;br /&gt; \frac{xy}{ax^2+ bxy+ cy^2}= \frac{sin(\theta)cos(\theta)}{acos^2(\theta)+ bsin(\theta)cos(\theta)+ csin^2(\theta)}.&lt;br /&gt; &lt;br /&gt; Notice that there is no &amp;quot;r&amp;quot; in that! This can have a limit as r-&amp;gt; 0 only if it does NOT depend on \theta- it is a constant. One obvious choice for a,b,c is a= c= 0, b= 1 but there may be others.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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