\begin{align*}

\lim\limits_{x\to \infty }\left(\displaystyle \frac{x^3+x+2}{x^2-1}\right) & =\lim\limits_{x\to \infty }\left(\displaystyle \frac{x^3+x+2}{x^2-1}\cdot \displaystyle \frac{\displaystyle \frac{1}{x^3}}{\displaystyle \frac{1}{x^3}}\right) \\
\\
&=\lim\limits_{x\to \infty }\left(\displaystyle \frac{\displaystyle \frac{x^3}{x^3}+\displaystyle \frac{x}{x^3}+\displaystyle \frac{2}{x^3}}{\displaystyle \frac{x^2}{x^3}-\displaystyle \frac{1}{x^3}}\right)\\
\\
&=\lim\limits_{x\to \infty }\left(\displaystyle \frac{1+\displaystyle \frac{1}{x^2}+\displaystyle \frac{2}{x^3}}{\displaystyle \frac{1}{x}-\displaystyle \frac{1}{x^3}}\right)\\
\\
&=\displaystyle \frac{1+0+0}{0-0}\\
\\
&=\infty \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

\begin{align*}
\lim\limits_{x\to \infty }\left(\frac{4x+5}{x^2+1}\right) & =\lim\limits_{x\to \infty }\left(\frac{4x+5}{x^2+1}\cdot \frac{\displaystyle\frac{1}{x^2}}{\displaystyle\frac{1}{x^2}}\right) \\
\\
&=\lim\limits_{x\to \infty }\left(\frac{\displaystyle\frac{4x}{x^2}+\displaystyle\frac{5}{x^2}}{\displaystyle\frac{x^2}{x^2}+\displaystyle\frac{1}{x^2}}\right)\\
\\
&=\lim\limits_{x\to \infty }\left(\frac{\displaystyle\frac{4}{x}+\displaystyle\frac{5}{x^2}}{1+\displaystyle\frac{1}{x^2}}\right) \\
\\
&=\displaystyle\frac{0+0}{1+0} \\
\\
&=0 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

\begin{align*}

\displaystyle \lim\limits_{x\to \infty }\left(\frac{3x^2+x+2}{x^3+8x+1}\right) & =\displaystyle \lim\limits_{x\to \infty }\left(\frac{3x^2+x+2}{x^3+8x+1}\cdot \frac{\displaystyle \frac{1}{x^3}}{\displaystyle \frac{1}{x^3}}\right)\\
\\
&=\lim\limits_{x\to \infty }\left(\displaystyle \frac{\displaystyle \frac{3x^2}{x^3}+\displaystyle \frac{x}{x^3}+\displaystyle \frac{2}{x^3}}{\displaystyle \frac{x^3}{x^3}+\frac{8x}{x^3}+\frac{1}{x^3}}\right)\\
\\
& =\lim\limits_{x\to \infty }\left(\displaystyle \frac{\displaystyle \frac{3}{x}+\frac{1}{x^2}+\displaystyle \frac{2}{x^3}}{1+\displaystyle \frac{8}{x^2}+\frac{1}{x^3}}\right)\\
\\
&=\frac{0+0+0}{1+0+0} \\
\\
&=0\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

\begin{align*}
\displaystyle \lim\limits_{x\to \infty }\left(\frac{6x^3+4x^2+5}{8x^3+7x-3}\right)&=\displaystyle \lim\limits_{x\to \infty }\left(\frac{6x^3+4x^2+5}{8x^3+7x-3}\cdot \displaystyle \frac{\displaystyle \frac{1}{x^3}}{\displaystyle \frac{1}{x^3}}\right) \\
\\
& =\displaystyle \lim\limits_{x\to \infty \:}\left(\displaystyle \frac{6+\displaystyle \frac{4}{x}+\displaystyle \frac{5}{x^3}}{8+\displaystyle \frac{7}{x^2}-\frac{3}{x^3}}\right)\\
\\
& =\displaystyle \frac{\lim\limits_{x\to \infty \:}\left(6+\displaystyle \frac{4}{x}+\frac{5}{x^3}\right)}{\lim\limits_{x\to \infty \:}\left(8+\displaystyle\frac{7}{x^2}-\displaystyle \frac{3}{x^3}\right)}\\
\\
& =\displaystyle \frac{6+0+0}{8+0-0}\\
\\
& =\displaystyle \frac{6}{8}\\
\\
& =\displaystyle \frac{3}{4} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

\begin{align*}
\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{f\left(x+2\right)-f\left(2\right)}{x}\right)&=\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\left(\left(x+2\right)^2-2\left(x+2\right)+3\right)-\left(2^2-2\cdot 2+3\right)}{x}\right)\\
\\
& =\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\left(\left(x+2\right)^2-2\left(x+2\right)+3\right)-\left(3\right)}{x}\right)\\
\\
&=\displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\left(\left(x+2\right)^2-2\left(x+2\right)\right)}{x}\right)\\
\\
\end{align*}

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

\begin{align*}
\displaystyle \lim\limits_{x\to 2}\left(\displaystyle \frac{f\left(x\right)-f\left(2\right)}{x-2}\right) & =\lim\limits_{x\to 2}\left(\displaystyle \frac{\left(x^2-2x+3\right)-\left(2^2-2\cdot 2+3\right)}{x-2}\right) \\
\\
& =\lim\limits_{x\to 2}\left(\displaystyle \frac{\left(x^2-2x+3\right)-3}{x-2}\right)\\
\\
& =\lim\limits_{x\to 2}\left(\displaystyle \frac{x^2-2x}{x-2}\right)\\
\\

\end{align*}

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

\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*}