Find the limit if exist: Lim 4xy+5yz+4xz/16x^2+25y^2+16z^2 (x,y,z)-(0,0,0)

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Find the limit if exist: Lim 4xy+5yz+4xz/16x^2+25y^2+16z^2 (x,y,z)-(0,0,0)

Mathematics
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Well, multivariate limits are always difficult because there are quite a few combination's of paths to take in order to solve it. Especially, when there are continuity problems like in this limit, i.e. discontinuous @ (x,y,z)=(0,0,0). Here's what I did: \[\lim_{(x,y,z) \rightarrow (0,0,0)}(4xy+5yz+4xz)/(16x^{2}+25y^{2}+16z^{2})\] If you try to go on the path of strictly the x, y, and z axes, meaning the other respective variables are 0, you get the limit = 0, for example: \[\lim_{(0,y,0) \rightarrow (0,0,0)}(4(0)(0)+5y(0)+4(0)(0))/16(0)+25y^{2}+16(0)\]\[\lim_{(0,y,0) \rightarrow (0,0,0)}0/25y^{2}=0\] However, if you go along the path y=x you'll get the following limit. \[\lim_{(x,y,z) \rightarrow (x,x,0)}(4xx+5x(0)+4x(0))/(16x^{2}+25x^{2}+16(0))\]\[\lim_{(x,y,z) \rightarrow (x,x,0)}(4x^{2})/(16x^{2}+25x^{2})=4x^{2}/x^{2}(16+25)=4/41\] Since you've produced a different value, other than 0, by using a path different from the previous paths the limit does not exist.
Well, multivariate limits are always difficult because there are quite a few combination's of paths to take in order to solve it. Especially, when there are continuity problems like in this limit, i.e. discontinuous @ (x,y,z)=(0,0,0). Here's what I did: lim(x,y,z)→(0,0,0)(4xy+5yz+4xz)/(16x2+25y2+16z2) If you try to go on the path of strictly the x, y, and z axes, meaning the other respective variables are 0, you get the limit = 0, for example: lim(0,y,0)→(0,0,0)(4(0)(0)+5y(0)+4(0)(0))/16(0)+25y2+16(0) lim(0,y,0)→(0,0,0)0/25y2=0 However, if you go along the path y=x you'll get the following limit. lim(x,y,z)→(x,x,0)(4xx+5x(0)+4x(0))/(16x2+25x2+16(0)) lim(x,y,z)→(x,x,0)(4x2)/(16x2+25x2)=4x2/x2(16+25)=4/41 Since you've produced a different value, other than 0, by using a path different from the previous paths the limit does not exist. The math didn't show very well above, this is a more succinct version.
Thanks !!!

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