\begin{align*}
\displaystyle \lim\limits_{x\to \pi }\left(\displaystyle \frac{\sin^2\left(x\right)}{1+\cos\left(x\right)}\right) & = \displaystyle \lim\limits_{x\to \pi }\left(\displaystyle \frac{1-\cos^2\left(x\right)}{1+\cos\left(x\right)}\right) \\
\\
& = \displaystyle \lim\limits_{x\to \pi }\left(\displaystyle \frac{\left(1+\cos\left(x\right)\right)\left(1-\cos\left(x\right)\right)}{1+\cos\left(x\right)}\right)\\
\\
& =\lim\limits_{x\to \pi }\left(1-\cos\left(x\right)\right)\\
\\
& =\left(1-\cos\left(\pi \right)\right)\\
\\
& =\left(1-\left(-1\right)\right)\\
\\
& =2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

\begin{align*}
\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\sin\left(x\right)\sin\left(2x\right)}{1-\cos\left(x\right)}\right) & =\lim\limits_{x\to 0}\left(\frac{\sin\left(x\right)\cdot 2\left(\sin\left(x\right) \cos\left(x\right)\right)}{1-\cos\left(x\right)}\right)\\
\\
& =\displaystyle  2\cdot \lim\limits_{x\to 0}\left(\frac{\sin^2\left(x\right)\cos\left(x\right)}{1-\cos\left(x\right)}\right)\\
\end{align*}

\begin{align*}
\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\sin\left(x\right)\sin\left(2x\right)}{1-\cos\left(x\right)}\right) & =2\cdot \lim\limits_{x\to 0}\left(\frac{\left(1-\cos^2\left(x\right)\right)\cos\left(x\right)}{1-\cos\left(x\right)}\right)\\
\\
& =2\cdot \lim\limits_{x\to 0}\left(\frac{\left(1+\cos\left(x\right)\right)\left(1-\cos\left(x\right)\right)\cos\left(x\right)}{1-\cos\left(x\right)}\right) \\
\\
& =2\cdot \lim\limits_{x\to 0}\left(\left(1+\cos\left(x\right)\right)\cos\left(x\right)\right) \\
\\
& = 2\cdot \left(\left(1+\cos\left(0\right)\right)\cos\left(0\right)\right) \\
\\
& =2\cdot \left(\left(1+1\right)\cdot 1\right) \\
\\
& =4 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

\begin{align*}
\lim\limits_{x\to 0}\left(\frac{1-\cos^2  x }{1+\cos x }\right) & =\frac{1-\cos^2\left(0\right)}{1+\cos\left(0\right)} \\
\\
& =\frac{1-1}{1+1}\\ 
\\
& = 0 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

\begin{align*}
\displaystyle \lim _{x\to 0}\left(\frac{\sin^3x}{\sin x-\tan x}\right) & =\lim _{x\to 0}\left(\frac{\sin^3x}{\sin x-\frac{\sin x}{\cos x}}\right) \\ 
\\
& =\lim _{x\to 0}\left(\frac{\sin^3 x}{\frac{\sin x \cos x - \sin x}{\cos x}}\right) \\
\\
& = \lim _{x\to 0}\left(\frac{\sin^3 x \cos x }{\sin x \cos x - \sin x}\right) \\
\\
&=\lim _{x\to \:0}\left(\frac{\sin^3 x \cos x}{\left(\sin x \right)\left(\cos x-1\right)}\right) \\
\\
& =\lim _{x\to 0}\left(\frac{\sin^2 x \cos x }{\left(\cos x -1\right)}\right) \\
\\
& =\lim _{x\to 0}\left(\frac{\left(1-\cos^2 x \right) \cdot\cos x }{-\left(1-\cos x \right)}\right) \\
\\
& =\lim\limits_{x\to 0}\left(\frac{\left(1+\cos x \right) \left(1-\cos x \right) \cdot\cos x }{-\left(1-\cos x \right)}\right) \\
\\
& =\lim _{x\to 0}\left(\frac{\left(1+\cos x \right) \cdot \cos\left(x\right)}{-1}\right) \\
\\
& =-1\cdot \lim _{x\to 0}\left(\left(1+\cos x\right)\cdot\cos x \right) \\
\\
& =-1\cdot \left(1+\cos 0 \right)\cdot \cos 0 \\
\\
& =-1\cdot \left(1+1\right)\cdot 1 \\
\\
& =-2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

\begin{align*}
\displaystyle \lim\limits_{x\to \:\frac{\pi \:}{4}}\left(\frac{\tan\:2x}{\sec\:2x}\right) & =\lim\limits_{x\to \frac{\pi }{4}}\left(\frac{\frac{\sin\:2x}{\cos\:2x}}{\frac{1}{\cos\:2x}}\right) \\
\\
& =\lim\limits_{x\to \:\frac{\pi \:}{4}}\left(\sin\:2x\right) \\
\\
& =\sin\left(2\cdot \frac{\pi }{4}\right) \\
\\
& =\sin\frac{\pi }{2} \\
\\
& =1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

\begin{align*}
\lim\limits_{x\to \:3}\left(\frac{\sqrt{x^2-9}}{x-3}\right) & =\lim\limits_{x\to 3}\left(\frac{\sqrt{x^2-9}}{x-3}\cdot \frac{\sqrt{x^2-9}}{\sqrt{x^2-9}}\right) \\
\\
& =\lim\limits_{x\to 3}\left(\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\sqrt{x^2-9}}\right) \\
\\
& =\lim\limits_{x\to 3}\left(\frac{x^2-9}{\left(x-3\right)\sqrt{x^2-9}}\right) \\
\\
& =\lim _{x\to 3}\left(\frac{x+3}{\sqrt{x^2-9}}\right) \\
\\
& =\frac{3+3}{\sqrt{3^2-9}} \\
\\
& =\frac{6}{0} \\
\\
& =\infty \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

\begin{align*}
\lim\limits_{x\to \:0}\left(\frac{1}{x}\left(\frac{1}{3}-\frac{1}{\sqrt{x+9}}\right)\right) & =\lim\limits_{x\to 0}\left(\frac{1\cdot \:\left(\frac{1}{3}-\frac{1}{\sqrt{x+9}}\right)}{x}\right) \\ \\
& =\lim\limits_{x\to 0}\left(\frac{\frac{\sqrt{x+9}-3}{3\sqrt{x+9}}}{x}\right) \\ \\
& =\lim\limits_{x\to 0}\left(\frac{\sqrt{x+9}-3}{3x\sqrt{x+9}}\right) \\ \\
& =\lim\limits_{x\to 0}\left(\frac{\sqrt{x+9}-3}{3x\sqrt{x+9}}\right)\cdot \frac{\sqrt{x+9}+3}{\sqrt{x+9}+3} \\ \\
& =\lim\limits_{x\to 0}\left(\frac{x}{3x\left(\sqrt{x+9}+3\right)\sqrt{x+9}}\right) \\ \\
& =\lim\limits_{x\to 0}\left(\frac{1}{3\left(\sqrt{x+9}+3\right)\sqrt{x+9}}\right) \\ \\
& =\frac{1}{3\left(\sqrt{0+9}+3\right)\sqrt{0+9}} \\ \\
& =\frac{1}{54} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

\begin{align*}
\begin{align*}

\lim\limits_{x\to \:3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right) & =\lim\limits_{x\to \:\:3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right)\cdot \frac{\sqrt{x-2}+\sqrt{4-x}}{\sqrt{x-2}+\sqrt{4-x}} \\

& =\lim\limits_{x\to 3}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{\left(\sqrt{x-2}-\sqrt{4-x}\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}\right] \\

& =\lim\limits_{x\to3}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{2x-6}\right]\\

& =\lim\limits_{x\to3}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{-x+4}\right)}{2\left(x-3\right)}\right] \\

& =\lim\limits_{x\to 3}\left[\frac{\sqrt{x-2}+\sqrt{4-x}}{2}\right]\\

& =\frac{\sqrt{3-2}+\sqrt{4-3}}{2}\\

& =1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}
\end{align*}

\begin{align*}
\lim\limits_{x\to 2}\left(\frac{x^3-8}{x^2-4\:}\right) & =\lim\limits_{x\to 2}\left[\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x-2\right)}\right] \\

&=\lim\limits_{x\to 2}\left[\frac{\left(x^2+2x+4\right)}{\left(x+2\right)}\right] \\

& =\frac{2^2+2\cdot 2+4}{2+2} \\

& =3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

\begin{align*}
\\
 \lim\limits_{x\to 4}\left(\frac{\frac{1}{x}-\frac{1}{4}}{x-4}\right)& =\lim\limits_{x\to 4}\left(\frac{\frac{4-x}{4x}}{x-4}\right)\\
\\
& =\lim\limits_{x\to 4}\frac{4-x}{4x\left(x-4\right)}\\
\\

&=\lim\limits_{x\to 4}\left(\frac{4-x}{-4x\left(4-x\right)}\right)\\
\\

& =\lim\limits_{x\to 4}-\frac{1}{4x}\\
\\

& =-\frac{1}{4\cdot 4}\\
\\

& =-\frac{1}{16} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\\
\end{align*}