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.

HINT: 
$$
 \begin{eqnarray}
 \frac{1}{x-4} \left ( \frac{8}{\sqrt{4+3x}}-2\right) &=&\frac{1}{x-4} \left ( \frac{8}{\sqrt{4+3x}}-2\right) \left ( \frac{8}{\sqrt{4+3x}}+2\right) \left ( \frac{8}{\sqrt{4+3x}}+2\right)^{-1} \\
   &=& \frac{1}{x-4} \left( \frac{64}{4+3x} -4 \right) \left ( \frac{8}{\sqrt{4+3x}}+2\right)^{-1} \\
   &=&  \frac{1}{x-4} \left( \frac{-12(x-4)}{4+3x} \right) \left ( \frac{8}{\sqrt{4+3x}}+2\right)^{-1} \\ &=& \left( \frac{-12}{4+3x} \right) \left ( \frac{8}{\sqrt{4+3x}}+2\right)^{-1}
\end{eqnarray}
$$
Can you find the limit now ?
