Question

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

Answer 1

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.

Answer 2

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.

Answer 3

Thanks !!!