\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{16tan(6x)}{11x^{3}}\rightarrow\frac{0}{0}\\&\frac{96tan(6x)^{2} + 96}{33x^{2}}\rightarrow\frac{96}{0}=\infty\\&lim_{x\rightarrow0-}\frac{16tan(6x)}{11x^{3}}=\infty\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-18tan(18x)}{7ln((x + 1)^{5})}\rightarrow\frac{0}{0}\\&\frac{- 324tan(18x)^{2} - 324}{\frac{35}{(x + 1)}}\rightarrow\frac{-324}{35}\\&lim_{x\rightarrow0-}\frac{-18tan(18x)}{7ln((x + 1)^{5})}=-\frac{324}{35}\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{10sin(4x)^{3}}{3x^{3}}\rightarrow\frac{0}{0}\\&\frac{120cos(4x)sin(4x)^{2}}{9x^{2}}\rightarrow\frac{0}{0}\\&\frac{960cos(4x)^{2}sin(4x) - 480sin(4x)^{3}}{18x}\rightarrow\frac{0}{0}\\&\frac{3840cos(4x)^{3} - 13440cos(4x)sin(4x)^{2}}{18}\rightarrow\frac{3840}{18}\\&lim_{x\rightarrow0-}\frac{10sin(4x)^{3}}{3x^{3}}=\frac{640}{3}\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-9sin(15\cdot \pi x)^{2}}{11\cdot (x - 1)^{2}}\rightarrow\frac{0}{0}\\&\frac{-270\cdot \pi cos(15\cdot \pi x)sin(15\cdot \pi x)}{22x - 22}\rightarrow\frac{0}{0}\\&\frac{4050\cdot \pi ^{2}sin(15\cdot \pi x)^{2} - 4050\cdot \pi ^{2}cos(15\cdot \pi x)^{2}}{22}\rightarrow\frac{-4050\cdot \pi ^{2}}{22}\\&lim_{x\rightarrow1+}\frac{-9sin(15\cdot \pi x)^{2}}{11\cdot (x - 1)^{2}}=-\frac{(2025\cdot \pi ^{2})}{11}\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-sin(5x)}{3cos(\frac{(\pi x)}{2}-\frac{ \pi }{2})}\rightarrow\frac{0}{0}\\&\frac{-5cos(5x)}{-\frac{(3\cdot \pi sin(\frac{(\pi x)}{2}-\frac{ \pi }{2}))}{2}}\rightarrow\frac{-5}{\frac{(3\cdot \pi )}{2}}\\&lim_{x\rightarrow0-}\frac{-sin(5x)}{3cos(\frac{(\pi x)}{2}-\frac{ \pi }{2})}=-\frac{10}{(3\cdot \pi )}\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{tan(6\cdot \pi x)^{2}}{(x - 1)^{3}}\rightarrow\frac{0}{0}\\&\frac{12\cdot \pi tan(6\cdot \pi x)\cdot (tan(6\cdot \pi x)^{2} + 1)}{3\cdot (x - 1)^{2}}\rightarrow\frac{0}{0}\\&\frac{72\cdot \pi ^{2}\cdot (tan(6\cdot \pi x)^{2} + 1)^{2} + 144\cdot \pi ^{2}tan(6\cdot \pi x)^{2}\cdot (tan(6\cdot \pi x)^{2} + 1)}{6x - 6}\rightarrow\frac{72\cdot \pi ^{2}}{0}=\infty\\&lim_{x\rightarrow1+}\frac{tan(6\cdot \pi x)^{2}}{(x - 1)^{3}}=\infty\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-31tan(2\cdot \pi x)^{3}}{27ln(x^{3})^{2}}\rightarrow\frac{0}{0}\\&\frac{-186\cdot \pi tan(2\cdot \pi x)^{2}\cdot (tan(2\cdot \pi x)^{2} + 1)}{\frac{(162ln(x^{3}))}{x}}\rightarrow\frac{0}{0}\\&\frac{- 744\cdot \pi ^{2}tan(2\cdot \pi x)\cdot (tan(2\cdot \pi x)^{2} + 1)^{2} - 744\cdot \pi ^{2}tan(2\cdot \pi x)^{3}\cdot (tan(2\cdot \pi x)^{2} + 1)}{\frac{486}{x^{2}}-\frac{ (162ln(x^{3}))}{x^{2}}}\rightarrow\frac{0}{486}=0\\&lim_{x\rightarrow1+}\frac{-31tan(2\cdot \pi x)^{3}}{27ln(x^{3})^{2}}=0\end{align*}\)

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

\(\displaystyle \begin{align*}\\&\frac{-sin(2\cdot \pi x)^{3}}{4ln(x^{8})^{3}}\rightarrow\frac{0}{0}\\&\frac{-6\cdot \pi cos(2\cdot \pi x)sin(2\cdot \pi x)^{2}}{\frac{(96ln(x^{8})^{2})}{x}}\rightarrow\frac{0}{0}\\&\frac{12\cdot \pi ^{2}sin(2\cdot \pi x)^{3} - 24\cdot \pi ^{2}cos(2\cdot \pi x)^{2}sin(2\cdot \pi x)}{\frac{(1536ln(x^{8}))}{x^{2}}-\frac{ (96ln(x^{8})^{2})}{x^{2}}}\rightarrow\frac{0}{0}\\&\frac{168\cdot \pi ^{3}cos(2\cdot \pi x)sin(2\cdot \pi x)^{2} - 48\cdot \pi ^{3}cos(2\cdot \pi x)^{3}}{\frac{12288}{x^{3}}-\frac{ (4608ln(x^{8}))}{x^{3}}+\frac{ (192ln(x^{8})^{2})}{x^{3}}}\rightarrow\frac{-48\cdot \pi ^{3}}{12288}\\&lim_{x\rightarrow1-}\frac{-sin(2\cdot \pi x)^{3}}{4ln(x^{8})^{3}}=-\frac{\pi ^{3}}{256}\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-7x^{2}}{37sin(15\cdot \pi x)^{3}}\rightarrow\frac{0}{0}\\&\frac{-14x}{1665\cdot \pi cos(15\cdot \pi x)sin(15\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\\&\frac{-14}{49950\cdot \pi ^{2}cos(15\cdot \pi x)^{2}sin(15\cdot \pi x) - 24975\cdot \pi ^{2}sin(15\cdot \pi x)^{3}}\rightarrow\frac{-14}{0}=\infty\\&lim_{x\rightarrow0-}\frac{-7x^{2}}{37sin(15\cdot \pi x)^{3}}=\infty\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-33sin(9\cdot \pi x)}{25ln(x^{3})^{2}}\rightarrow\frac{0}{0}\\&\frac{-297\cdot \pi cos(9\cdot \pi x)}{\frac{(150ln(x^{3}))}{x}}\rightarrow\frac{297\cdot \pi}{0}=-\infty\\&lim_{x\rightarrow1-}\frac{-33sin(9\cdot \pi x)}{25ln(x^{3})^{2}}=-\infty\end{align*}\)