\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*}

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

& =\lim\limits_{x\to 8}\:\frac{1}{\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}\\
\\

& =\frac{1}{\left(\sqrt[3]{8^2}+2\sqrt[3]{8}+4\right)}\\
\\

& =\frac{1}{4+4+4}\\
\\

& =\frac{1}{12} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\\
\end{align*}

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

\begin{align*}
\\
\lim\limits_{x\to 0}\:\frac{\sqrt{x+16}-4}{x} & =\lim\limits_{x\to 0}\:\frac{\sqrt{x+16}-4}{x}\cdot \frac{\sqrt{x+16}+4}{\sqrt{x+16}+4}\\
\\
 &  =\lim\limits_{x\to 0}\:\frac{\left(x+16\right)-4^2}{x\left(\sqrt{x+16}+4\right)}\\
\\
 & =\lim\limits_{x\to 0}\:\frac{x+16-16}{x\left(\sqrt{x+16}+4\right)}\\
\\
 &  =\lim\limits_{x\to 0}\:\frac{x}{x\left(\sqrt{x+16}+4\right)}\\
\\
 &  =\lim\limits_{x\to 0}\:\frac{1}{\sqrt{x+16}+4}\\
\\
 &  =\:\frac{1}{\sqrt{0+16}+4}\\
\\
 &  =\:\frac{1}{4+4}\\
\\
 &  =\:\frac{1}{8} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\\
\end{align*}

\begin{align*}

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

\begin{align*}

\lim\limits_{x\to 2}\left(\frac{x^3-x^2-x-2}{2x^3-5x^2+5x-6}\right)&=\lim\limits_{x\to 2}\left(\frac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-2\right)\left(2x^2-x+3\right)}\right)\\
\\
& =\lim\limits_{x\to 2}\left(\frac{x^2+x+1}{2x^2-x+3}\right)\\
\\
& =\frac{2^2+2+1}{2\left(2\right)^2-2+3}\\
\\
& =\frac{4+2+1}{8-2+3}\\
\\
& =\frac{7}{9} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\\
\end{align*}