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Find values for which the limit exists... 
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#1
Oct1605, 07:50 AM

P: 368

Hi,
I'm given this problem: Find conditions for variables a, b, c so that the limit [tex] \lim_{[x,y] \rightarrow [0,0]} \frac{xy}{ax^2 + bxy + cy^2} [/tex] exists. What I have only found so far is that for all variables nonzero 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. 


#2
Oct1605, 08:35 AM

PF Gold
P: 973

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.



#3
Oct1605, 08:42 AM

PF Gold
P: 973

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?



#4
Oct1605, 02:45 PM

P: 368

Find values for which the limit exists...
Anyway, it is just the result of guessing method, is there any more exact approach to solve this? 


#5
Oct1605, 03:13 PM

Math
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Thanks
PF Gold
P: 39,533

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,
[itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex] so [itex]xy= r^2 cos(\theta)sin(\theta)[/itex], [tex]x^{2}= r^2 cos^2(\theta)[/itex], and [tex]y^2= r^2 sin^2(\theta)[/itex]. Of course, then [itex]ax^2+ bxy+ cy^2= ar^2cos^2(\theta)+ br^2sin( theta)cos(\theta)+ cr^2sin^2(\theta)[/itex] so that [itex]ax^2+ bxy+ cy^2= r^2(acos^2(\theta)+ bsin( theta)cos(\theta)+ csin^2(\theta)[/itex]. That means that [tex]\frac{xy}{ax^2+ bxy+ cy^2}= \frac{sin(\theta)cos(\theta)}{acos^2(\theta)+ bsin(\theta)cos(\theta)+ csin^2(\theta)}[/tex]. Notice that there is no "r" in that! This can have a limit as r> 0 only if it does NOT depend on [itex]\theta[/itex] it is a constant. One obvious choice for a,b,c is a= c= 0, b= 1 but there may be others. 


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