Find the limit, if it exists. (If an answer does not exist, enter DNE.)

lim (x, y, z)→(0, 0, 0)

(xy+2yz^2+3xz^2) / (x^2+y^2+z^4)

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\lim _{(x, y, z) \rightarrow(0,0) 0)} \frac{x y+2 y z^{2}+3 x z^{2}}{x^{2}+y^{2}+24} .choose \quad y=m x, \quad y=m \cdot \sqrt{x}.\begin{array}{l}3=m v_{x} \\\end{array}\begin{array}{l}\begin{array}{l}=\lim _{x \rightarrow 0} \frac{x^{2} m+2 m^{3} x^{2}+3 m^{2} x^{2}}{x^{2}\left(1+m^{2}+m^{4}\right)} \\y=m x \\z=m \sqrt{x}\end{array} \\\begin{array}{l}=\lim _{x \rightarrow 0} x / \frac{\left(m+2 m^{3}+3 m^{2}\right)}{y=m x} x^{2}\left(1+m^{2}+m^{4}\right) \\z=m \sqrt{x}\end{array} \\\end{array}z=m \sqrt{n}.lyou have any doubt comment me, iwill replay. Don't forget ratting. Thanking you ...