\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(20(0)^{4}y^{2})}{(4(0)^{2}y^{4} + 14(0)^{3}y^{3} + 10(0)y^{5} - (0)^{5}y + (0)^{6} + y^{6})}=\frac{0}{y^{6}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(20x^{4}(x)^{2})}{(4x^{2}(x)^{4} + 14x^{3}(x)^{3} + 10x(x)^{5} - x^{5}(x) + x^{6} + (x)^{6})}=-\frac{20}{29}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{((0)^{2}y^{3} + 11(0)^{3}y^{2} - 12(0)^{5})}{(4(0)^{5} + 3y^{5})}=\frac{0}{3y^{5}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{3y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(x^{2}(x)^{3} + 11x^{3}(x)^{2} - 12x^{5})}{(4x^{5} + 3(x)^{5})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(5(0)^{4}y^{2} - 20(0)^{6})}{(19(0)^{4}y^{2} - 5(0)^{2}y^{4} + 7(0)^{5}y + (0)^{6} + y^{6})}=\frac{0}{y^{6}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(5x^{4}(x)^{2} - 20x^{6})}{(19x^{4}(x)^{2} - 5x^{2}(x)^{4} + 7x^{5}(x) + x^{6} + (x)^{6})}=\frac{15}{23}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(13(0)^{2}y^{2} - 13(0)y^{3})}{(13(0)^{4} - 19(0)^{3}y + y^{4})}=\frac{0}{y^{4}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(13x^{2}(x)^{2} - 13x(x)^{3})}{(13x^{4} - 19x^{3}(x) + (x)^{4})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(28(0)y^{4})}{(22(0)^{2}y^{3} + 2(0)y^{4} - 5(0)^{4}y + (0)^{5} + 5y^{5})}=\frac{0}{5y^{5}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{5y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(28x(x)^{4})}{(22x^{2}(x)^{3} + 2x(x)^{4} - 5x^{4}(x) + x^{5} + 5(x)^{5})}=\frac{28}{25}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(11(0)^{3}y^{3} - 20(0)^{5}y)}{(4(0)^{6} - 20(0)^{4}y^{2} + 2y^{6})}=\frac{0}{2y^{6}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{2y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(11x^{3}(x)^{3} - 20x^{5}(x))}{(4x^{6} - 20x^{4}(x)^{2} + 2(x)^{6})}=-\frac{9}{14}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(15(0)^{4}y - 7(0)^{3}y^{2} + 18(0)^{5})}{(34(0)^{4}y + 3(0)^{5} + 5y^{5})}=\frac{0}{5y^{5}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{5y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}\frac{(15x^{4}(x) - 7x^{3}(x)^{2} + 18x^{5})}{(34x^{4}(x) + 3x^{5} + 5(x)^{5})}=\frac{13}{21}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(8(0)^{2}y^{4})}{(19(0)^{2}y^{4} + 13(0)y^{5} + 4(0)^{6} - 2y^{6})}=\frac{0}{2y^{6}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{2y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(8x^{2}(x)^{4})}{(19x^{2}(x)^{4} + 13x(x)^{5} + 4x^{6} - 2(x)^{6})}=-\frac{4}{17}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(5(0)y^{3} - 17(0)^{3}y + 12(0)^{4})}{(2(0)^{4} + 3y^{4})}=\frac{0}{3y^{4}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{3y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(5x(x)^{3} - 17x^{3}(x) + 12x^{4})}{(2x^{4} + 3(x)^{4})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}\)

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