\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(16(0)^{3}y^{3})}{(7(0)^{4}y^{2} + 15(0)^{5}y + 5(0)^{6} + 4y^{6})}=\frac{0}{4y^{6}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{4y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}\frac{(16x^{3}(x)^{3})}{(7x^{4}(x)^{2} + 15x^{5}(x) + 5x^{6} + 4(x)^{6})}=\frac{16}{31}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(2(0)^{3}y^{3} - 3(0)^{6} + y^{6})}{(2(0)^{6} + 4y^{6})}=\frac{-y^{6}}{4y^{6}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{-y^{6}}{4y^{6}}=-\frac{1}{4}\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(2x^{3}(x)^{3} - 3x^{6} + (x)^{6})}{(2x^{6} + 4(x)^{6})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(8(0)^{2}y^{3} - 16(0)^{3}y^{2} + 8(0)y^{4})}{(2(0)^{5} + 4y^{5})}=\frac{0}{4y^{5}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{4y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(8x^{2}(x)^{3} - 16x^{3}(x)^{2} + 8x(x)^{4})}{(2x^{5} + 4(x)^{5})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(7(0)^{2}y - 11(0)^{3} + 4y^{3})}{((0)^{3} + 5y^{3})}=\frac{-4y^{3}}{5y^{3}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{-4y^{3}}{5y^{3}}=-\frac{4}{5}\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(7x^{2}(x) - 11x^{3} + 4(x)^{3})}{(x^{3} + 5(x)^{3})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(14(0)^{3}y - 11(0)^{4})}{(17(0)y^{3} + (0)^{4} + 5y^{4})}=\frac{0}{5y^{4}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{5y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(14x^{3}(x) - 11x^{4})}{(17x(x)^{3} + x^{4} + 5(x)^{4})}=-\frac{3}{23}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(4(0)^{2}y - 11(0)y^{2} + 7(0)^{3})}{(5(0)^{3} + 4y^{3})}=\frac{0}{4y^{3}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{4y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(4x^{2}(x) - 11x(x)^{2} + 7x^{3})}{(5x^{3} + 4(x)^{3})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(8(0)y^{4} - 15(0)^{2}y^{3} + 7(0)^{5})}{(5(0)^{5} + y^{5})}=\frac{0}{y^{5}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(8x(x)^{4} - 15x^{2}(x)^{3} + 7x^{5})}{(5x^{5} + (x)^{5})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(15(0)^{5}y - 20(0)^{3}y^{3} + 10(0)^{6})}{(2(0)^{6} - (0)^{4}y^{2} + y^{6})}=\frac{0}{y^{6}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(15x^{5}(x) - 20x^{3}(x)^{3} + 10x^{6})}{(2x^{6} - x^{4}(x)^{2} + (x)^{6})}=\frac{5}{2}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

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

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(14(0)y^{5} - 18(0)^{2}y^{4} + 4(0)^{6})}{(3(0)^{6} + 2y^{6})}=\frac{0}{2y^{6}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{2y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(14x(x)^{5} - 18x^{2}(x)^{4} + 4x^{6})}{(3x^{6} + 2(x)^{6})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}\)