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.

The definition of the derivative, $\lim\limits_{x\to 4} \dfrac{f(x)-f(4)}{x-4}$ will always give you the indeterminate form $0/0$ if you plug in the number that $x$ is approaching.  I.e. you get $\dfrac{f(4)-f(4)}{4-4}$.
So when you see
$$\lim_{ x \to 4} \frac{1}{x-4} \cdot \left ( \frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{3+3\cdot4}}\right)$$
what you want is to find a factor of $x-4$ in the numerator that will cancel the $x-4$ in the denominator.  To do that, you want to write that difference of two fractions as just one fraction.  For that you use a common denominator:
$$
\frac{8}{\sqrt{4+3x}}-\frac{8}{\sqrt{3+3\cdot4}} = \frac{8}{\sqrt{4+3x}} - 2 = \frac{8}{\sqrt{4+3x}} - \frac{2\sqrt{4+3x}}{\sqrt{4+3x}} = \frac{8-2\sqrt{4+3x}}{\sqrt{4+3x}}
$$
When you plug $4$ into this, then as expected, you get $0$.  Now rationalize the numerator:
$$
\frac{8-2\sqrt{4+3x}}{\sqrt{4+3x}} = \frac{8-2\sqrt{4+3x}}{\sqrt{4+3x}} \cdot \frac{8+2\sqrt{4+3x}}{8+2\sqrt{4+3x}} = \frac{64-4(4+3x)}{\sqrt{4+3x}(8+2\sqrt{4+3x})}
$$
$$
= \frac{-12(x-4)}{\sqrt{4+3x}(8+2\sqrt{4+3x})}
$$
When you multiply this by $\dfrac{1}{x-4}$, you get a cancellation, and then you can find the limit just by substituting $4$ for $x$.
