\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-9tan(18x)}{ln((x + 1)^{4})^{3}}\rightarrow\frac{0}{0}\\&\frac{- 162tan(18x)^{2} - 162}{\frac{(12ln((x + 1)^{4})^{2})}{(x + 1)}}\rightarrow\frac{-162}{0}=-\infty\\&lim_{x\rightarrow0-}\frac{-9tan(18x)}{ln((x + 1)^{4})^{3}}=-\infty\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-5ln((x + 1)^{8})^{3}}{16x}\rightarrow\frac{0}{0}\\&\frac{-\frac{(120ln((x + 1)^{8})^{2})}{(x + 1)}}{16}\rightarrow\frac{0}{16}=0\\&lim_{x\rightarrow0+}\frac{-5ln((x + 1)^{8})^{3}}{16x}=0\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{4sin(3\cdot \pi x)^{3}}{9\cdot (x - 1)^{2}}\rightarrow\frac{0}{0}\\&\frac{36\cdot \pi cos(3\cdot \pi x)sin(3\cdot \pi x)^{2}}{18x - 18}\rightarrow\frac{0}{0}\\&\frac{216\cdot \pi ^{2}cos(3\cdot \pi x)^{2}sin(3\cdot \pi x) - 108\cdot \pi ^{2}sin(3\cdot \pi x)^{3}}{18}\rightarrow\frac{0}{18}=0\\&lim_{x\rightarrow1+}\frac{4sin(3\cdot \pi x)^{3}}{9\cdot (x - 1)^{2}}=0\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-3\cdot (x - 1)^{3}}{2ln(x^{8})^{3}}\rightarrow\frac{0}{0}\\&\frac{-9\cdot (x - 1)^{2}}{\frac{(48ln(x^{8})^{2})}{x}}\rightarrow\frac{0}{0}\\&\frac{18 - 18x}{\frac{(768ln(x^{8}))}{x^{2}}-\frac{ (48ln(x^{8})^{2})}{x^{2}}}\rightarrow\frac{0}{0}\\&\frac{-18}{\frac{6144}{x^{3}}-\frac{ (2304ln(x^{8}))}{x^{3}}+\frac{ (96ln(x^{8})^{2})}{x^{3}}}\rightarrow\frac{-18}{6144}\\&lim_{x\rightarrow1+}\frac{-3\cdot (x - 1)^{3}}{2ln(x^{8})^{3}}=-\frac{3}{1024}\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-sin(10\cdot \pi x)^{2}}{18ln(x^{5})^{3}}\rightarrow\frac{0}{0}\\&\frac{-20\cdot \pi cos(10\cdot \pi x)sin(10\cdot \pi x)}{\frac{(270ln(x^{5})^{2})}{x}}\rightarrow\frac{0}{0}\\&\frac{200\cdot \pi ^{2}sin(10\cdot \pi x)^{2} - 200\cdot \pi ^{2}cos(10\cdot \pi x)^{2}}{\frac{(2700ln(x^{5}))}{x^{2}}-\frac{ (270ln(x^{5})^{2})}{x^{2}}}\rightarrow\frac{-200\cdot \pi ^{2}}{0}=-\infty\\&lim_{x\rightarrow1+}\frac{-sin(10\cdot \pi x)^{2}}{18ln(x^{5})^{3}}=-\infty\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-5ln(x^{8})^{2}}{14\cdot (x - 1)^{2}}\rightarrow\frac{0}{0}\\&\frac{-\frac{(80ln(x^{8}))}{x}}{28x - 28}\rightarrow\frac{0}{0}\\&\frac{\frac{(80ln(x^{8}))}{x^{2}}-\frac{ 640}{x^{2}}}{28}\rightarrow\frac{-640}{28}\\&lim_{x\rightarrow1-}\frac{-5ln(x^{8})^{2}}{14\cdot (x - 1)^{2}}=-\frac{160}{7}\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{12sin(12\cdot \pi x)^{2}}{(x - 1)^{2}}\rightarrow\frac{0}{0}\\&\frac{288\cdot \pi cos(12\cdot \pi x)sin(12\cdot \pi x)}{2x - 2}\rightarrow\frac{0}{0}\\&\frac{3456\cdot \pi ^{2}cos(12\cdot \pi x)^{2} - 3456\cdot \pi ^{2}sin(12\cdot \pi x)^{2}}{2}\rightarrow\frac{3456\cdot \pi ^{2}}{2}\\&lim_{x\rightarrow1-}\frac{12sin(12\cdot \pi x)^{2}}{(x - 1)^{2}}=1728\cdot \pi ^{2}\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{13tan(11\cdot \pi x)}{8x^{2}}\rightarrow\frac{0}{0}\\&\frac{143\cdot \pi \cdot (tan(11\cdot \pi x)^{2} + 1)}{16x}\rightarrow\frac{143\cdot \pi}{0}=-\infty\\&lim_{x\rightarrow0-}\frac{13tan(11\cdot \pi x)}{8x^{2}}=-\infty\end{align*}\)

\(\displaystyle \begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{18tan(5\cdot \pi x)^{2}}{18sin(3\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\\&\frac{180\cdot \pi tan(5\cdot \pi x)\cdot (tan(5\cdot \pi x)^{2} + 1)}{108\cdot \pi cos(3\cdot \pi x)sin(3\cdot \pi x)}\rightarrow\frac{0}{0}\\&\frac{900\cdot \pi ^{2}\cdot (tan(5\cdot \pi x)^{2} + 1)^{2} + 1800\cdot \pi ^{2}tan(5\cdot \pi x)^{2}\cdot (tan(5\cdot \pi x)^{2} + 1)}{324\cdot \pi ^{2}cos(3\cdot \pi x)^{2} - 324\cdot \pi ^{2}sin(3\cdot \pi x)^{2}}\rightarrow\frac{900\cdot \pi ^{2}}{324\cdot \pi ^{2}}\\&lim_{x\rightarrow0-}\frac{18tan(5\cdot \pi x)^{2}}{18sin(3\cdot \pi x)^{2}}=\frac{25}{9}\end{align*}\)

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

\(\displaystyle \begin{align*}\\&\frac{-13sin(18\cdot \pi x)^{3}}{12ln(x^{8})^{3}}\rightarrow\frac{0}{0}\\&\frac{-702\cdot \pi cos(18\cdot \pi x)sin(18\cdot \pi x)^{2}}{\frac{(288ln(x^{8})^{2})}{x}}\rightarrow\frac{0}{0}\\&\frac{12636\cdot \pi ^{2}sin(18\cdot \pi x)^{3} - 25272\cdot \pi ^{2}cos(18\cdot \pi x)^{2}sin(18\cdot \pi x)}{\frac{(4608ln(x^{8}))}{x^{2}}-\frac{ (288ln(x^{8})^{2})}{x^{2}}}\rightarrow\frac{0}{0}\\&\frac{1592136\cdot \pi ^{3}cos(18\cdot \pi x)sin(18\cdot \pi x)^{2} - 454896\cdot \pi ^{3}cos(18\cdot \pi x)^{3}}{\frac{36864}{x^{3}}-\frac{ (13824ln(x^{8}))}{x^{3}}+\frac{ (576ln(x^{8})^{2})}{x^{3}}}\rightarrow\frac{-454896\cdot \pi ^{3}}{36864}\\&lim_{x\rightarrow1-}\frac{-13sin(18\cdot \pi x)^{3}}{12ln(x^{8})^{3}}=-\frac{(3159\cdot \pi ^{3})}{256}\end{align*}\)