\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{f\left(9+x\right)-f\left(9\right)}{x}\right)=\lim\limits_{x\to 0}\left(\displaystyle \frac{\sqrt{9+x}-\sqrt{9}}{x}\right)

\begin{align*}
& =\lim\limits_{x\to 0}\left(\displaystyle \frac{\sqrt{9+x}-3}{x}\cdot \displaystyle \frac{\sqrt{9+x}+3}{\sqrt{9+x}+3}\right)\\
\\
& =\lim\limits_{x\to 0}\left(\displaystyle \frac{9+x-9}{x\left(\sqrt{9+x}+3\right)}\right)\\
\\
& =\lim\limits_{x\to 0}\left(\displaystyle \frac{x}{x\left(\sqrt{9+x}+3\right)}\right)\\
\\
& =\lim\limits_{x\to 0}\left(\displaystyle \frac{1}{\left(\sqrt{9+x}+3\right)}\right)\\
\\
& =\left(\displaystyle \frac{1}{\left(\sqrt{9+0}+3\right)}\right)\\
\\
& =\displaystyle \frac{1}{6} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

\displaystyle \lim\limits_{x\to 4}\left(\displaystyle \frac{f\left(x\right)-f\left(4\right)}{x-4}\right)=\lim\limits_{x\to 4}\left(\displaystyle \frac{\sqrt{x}-\sqrt{4}}{x-4}\right)

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

\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*}