\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(5(0)y^{2} + 13y^{3})}{(4(0)y^{2} + 25(0)^{3} - 2y^{3})}=\frac{-13y^{3}}{2y^{3}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{-13y^{3}}{2y^{3}}=-\frac{13}{2}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(5x(x)^{2} + 13(x)^{3})}{(4x(x)^{2} + 25x^{3} - 2(x)^{3})}=\frac{2}{3}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(5(0)y^{2} - 17(0)^{2}y + 6y^{3})}{(17(0)^{3} - 16y^{3})}=\frac{6y^{3}}{16y^{3}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{6y^{3}}{16y^{3}}=\frac{3}{8}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(5x(x)^{2} - 17x^{2}(x) + 6(x)^{3})}{(17x^{3} - 16(x)^{3})}=6\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(15(0)^{3}y^{2} - 3(0)^{4}y)}{(18(0)^{4}y - 20(0)^{2}y^{3} + 10(0)^{5} + 22y^{5})}=\frac{0}{22y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{22y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(15x^{3}(x)^{2} - 3x^{4}(x))}{(18x^{4}(x) - 20x^{2}(x)^{3} + 10x^{5} + 22(x)^{5})}=\frac{2}{5}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(19(0)^{5} + 12y^{5})}{(35(0)^{2}y^{3} + 16(0)y^{4} - 2(0)^{5} - 15y^{5})}=\frac{12y^{5}}{15y^{5}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{12y^{5}}{15y^{5}}=\frac{4}{5}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(19x^{5} + 12(x)^{5})}{(35x^{2}(x)^{3} + 16x(x)^{4} - 2x^{5} - 15(x)^{5})}=-\frac{31}{34}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(18(0)^{5})}{(18(0)^{2}y^{3} - 14(0)^{4}y + 15(0)^{5} + 2y^{5})}=\frac{0}{2y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{2y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(18x^{5})}{(18x^{2}(x)^{3} - 14x^{4}(x) + 15x^{5} + 2(x)^{5})}=\frac{6}{7}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(14(0)^{6})}{(16(0)^{3}y^{3} + 19(0)^{6} + 3y^{6})}=\frac{0}{3y^{6}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{3y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(14x^{6})}{(16x^{3}(x)^{3} + 19x^{6} + 3(x)^{6})}=\frac{7}{19}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{((0)y^{2} + 5(0)^{2}y - 6(0)^{3})}{((0)^{3} + 3y^{3})}=\frac{0}{3y^{3}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{3y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(x(x)^{2} + 5x^{2}(x) - 6x^{3})}{(x^{3} + 3(x)^{3})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(3(0)^{2}y^{2} - 12(0)y^{3} + 9(0)^{4})}{(3(0)^{4} + 2y^{4})}=\frac{0}{2y^{4}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{2y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(3x^{2}(x)^{2} - 12x(x)^{3} + 9x^{4})}{(3x^{4} + 2(x)^{4})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(11(0)y^{2} + 8(0)^{2}y)}{(11(0)^{2}y + 12(0)^{3} - 4y^{3})}=\frac{0}{4y^{3}}\\&\lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{4y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(11x(x)^{2} + 8x^{2}(x))}{(11x^{2}(x) + 12x^{3} - 4(x)^{3})}=-1\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(6(0)y^{3} + 5(0)^{4})}{(2(0)^{4} + 12y^{4})}=\frac{0}{12y^{4}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{12y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(6x(x)^{3} + 5x^{4})}{(2x^{4} + 12(x)^{4})}=\frac{11}{14}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)