Find values for which the limit exists

[twoflower]

Hi,
I'm given this problem:

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

$$\lim_{[x,y] \rightarrow [0,0]} \frac{xy}{ax^2 + bxy + cy^2}$$

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.

 

[Lisa!]

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.

 

[Lisa!]

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?

 

[twoflower]

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?

 

[HallsofIvy]

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
$xy= r^2 cos(\theta)sin(\theta)$, [tex]x²= r^2 cos^2(\theta)[/itex], and [tex]y^2= r^2 sin^2(\theta)[/itex].
Of course, then $ax^2+ bxy+ cy^2= ar^2cos^2(\theta)+ br^2sin( theta)cos(\theta)+ cr^2sin^2(\theta)$ so that
$ax^2+ bxy+ cy^2= r^2(acos^2(\theta)+ bsin( theta)cos(\theta)+ csin^2(\theta)$.

That means that
$$\frac{xy}{ax^2+ bxy+ cy^2}= \frac{sin(\theta)cos(\theta)}{acos^2(\theta)+ bsin(\theta)cos(\theta)+ csin^2(\theta)}$$.

Notice that there is no "r" in that! This can have a limit as r-> 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.