\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{49sin(19\cdot \pi x)^{2}}{22cos(\frac{(17\cdot \pi x)}{2})^{2}}\rightarrow\frac{0}{0}\\&\frac{1862\cdot \pi cos(19\cdot \pi x)sin(19\cdot \pi x)}{-374\cdot \pi cos(\frac{(17\cdot \pi x)}{2})sin(\frac{(17\cdot \pi x)}{2})}\rightarrow\frac{0}{0}\\&\frac{35378\cdot \pi ^{2}cos(19\cdot \pi x)^{2} - 35378\cdot \pi ^{2}sin(19\cdot \pi x)^{2}}{3179\cdot \pi ^{2}sin(\frac{(17\cdot \pi x)}{2})^{2} - 3179\cdot \pi ^{2}cos(\frac{(17\cdot \pi x)}{2})^{2}}\rightarrow\frac{35378\cdot \pi ^{2}}{3179\cdot \pi ^{2}}\\&lim_{x\rightarrow1+}\frac{49sin(19\cdot \pi x)^{2}}{22cos(\frac{(17\cdot \pi x)}{2})^{2}}=\frac{35378}{3179}\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-39x^{3}}{20sin(7\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\\&\frac{-117x^{2}}{280\cdot \pi cos(7\cdot \pi x)sin(7\cdot \pi x)}\rightarrow\frac{0}{0}\\&\frac{-234x}{1960\cdot \pi ^{2}cos(7\cdot \pi x)^{2} - 1960\cdot \pi ^{2}sin(7\cdot \pi x)^{2}}\rightarrow\frac{0}{1960\cdot \pi ^{2}}=0\\&lim_{x\rightarrow0-}\frac{-39x^{3}}{20sin(7\cdot \pi x)^{2}}=0\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{x^{2}}{sin(13\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\\&\frac{2x}{26\cdot \pi cos(13\cdot \pi x)sin(13\cdot \pi x)}\rightarrow\frac{0}{0}\\&\frac{2}{338\cdot \pi ^{2}cos(13\cdot \pi x)^{2} - 338\cdot \pi ^{2}sin(13\cdot \pi x)^{2}}\rightarrow\frac{2}{338\cdot \pi ^{2}}\\&lim_{x\rightarrow0-}\frac{x^{2}}{sin(13\cdot \pi x)^{2}}=\frac{1}{(169\cdot \pi ^{2})}\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-5x}{8sin(5\cdot \pi x)^{3}}\rightarrow\frac{0}{0}\\&\frac{-5}{120\cdot \pi cos(5\cdot \pi x)sin(5\cdot \pi x)^{2}}\rightarrow\frac{-5}{0}=-\infty\\&lim_{x\rightarrow0-}\frac{-5x}{8sin(5\cdot \pi x)^{3}}=-\infty\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{9sin(6x)^{2}}{7x^{2}}\rightarrow\frac{0}{0}\\&\frac{108cos(6x)sin(6x)}{14x}\rightarrow\frac{0}{0}\\&\frac{648cos(6x)^{2} - 648sin(6x)^{2}}{14}\rightarrow\frac{648}{14}\\&lim_{x\rightarrow0-}\frac{9sin(6x)^{2}}{7x^{2}}=\frac{324}{7}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(4(0)^{3}y^{3} + (0)y^{5} - 7(0)^{5}y)}{(14(0)^{6} - 27(0)^{3}y^{3} + y^{6})}=\frac{0}{y^{6}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(4x^{3}(x)^{3} + x(x)^{5} - 7x^{5}(x))}{(14x^{6} - 27x^{3}(x)^{3} + (x)^{6})}=-\frac{1}{6}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(7(0)y^{5} - 14(0)^{5}y)}{(8(0)y^{5} + 5(0)^{5}y + 16(0)^{6} - 3y^{6})}=\frac{0}{3y^{6}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{3y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}\frac{(7x(x)^{5} - 14x^{5}(x))}{(8x(x)^{5} + 5x^{5}(x) + 16x^{6} - 3(x)^{6})}=-\frac{7}{26}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(13(0)y^{4})}{(9(0)^{2}y^{3} + 13(0)^{5} - y^{5})}=\frac{0}{y^{5}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(13x(x)^{4})}{(9x^{2}(x)^{3} + 13x^{5} - (x)^{5})}=\frac{13}{21}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)

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

\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(20(0)^{3}y^{3} + 13(0)^{4}y^{2})}{(15(0)^{2}y^{4} + (0)y^{5} + 5(0)^{6} + 4y^{6})}=\frac{0}{4y^{6}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{4y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(20x^{3}(x)^{3} + 13x^{4}(x)^{2})}{(15x^{2}(x)^{4} + x(x)^{5} + 5x^{6} + 4(x)^{6})}=\frac{33}{25}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)\(\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(20(0)^{3}y^{3} + 13(0)^{4}y^{2})}{(15(0)^{2}y^{4} + (0)y^{5} + 5(0)^{6} + 4y^{6})}=\frac{0}{4y^{6}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{4y^{6}}=0\\&\text{We can make a similar substitution for }x=y:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(20x^{3}(x)^{3} + 13x^{4}(x)^{2})}{(15x^{2}(x)^{4} + x(x)^{5} + 5x^{6} + 4(x)^{6})}=\frac{33}{25}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}\)