Finding the slope of the tangent line to $\frac{8}{\sqrt{4+3x}}$ at $(4,2)$ In order to find the slope of the tangent line at the point $(4,2)$ belong to the function $\frac{8}{\sqrt{4+3x}}$, I choose the derivative at a given point formula.
$\begin{align*} 
\lim_{x \to 4} \frac{f(x)-f(4)}{x-4} &= 
\lim_{x \mapsto 4} \frac{1}{x-4} \cdot \left (\frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{4+3 \cdot 4}} \right ) 
\\ \\ & = \lim_{x \to 4} \frac{1}{x-4} \cdot \left ( \frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{16}} \right ) \\ \\ & = \lim_{ x \to 4} \frac{1}{x-4} \cdot \left ( \frac{8}{\sqrt{4+3x}}-2\right) 
\end{align*}$
But now I can't figure it out, how to end this limit.
I know that the derivative formula for this function is $-\frac{12}{(4+3x)\sqrt{4+3x}}$.
Thanks for the help.

Use the L'Hopital method, several equations becomes easy to solve

$$\lim_{x \to 4}\frac{f(x)-f(4)}{x-4}=\lim_{x \to 4}\frac{f'(x)-0}{1-0}=\lim_{x \to 4}f'(x)$$

Where
$f(x)=\frac{8}{\sqrt{3 x+4}}$ and $f'(x)$, the derivative of $f(x)$ is defined by

$f'(x)=-\frac{12}{(3 x+4)^{3/2}}$

The final equation results, just do the final calculus:
$$\lim_{x \to 4}\frac{f(x)-f(4)}{x-4}=\lim_{x \to 4}-\frac{12}{(3 x+4)^{3/2}}$$
