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.

All these answers are simple fact (except using L' Hospital, which is the best approach though) that you have to multiply by conjugate surds, which is greatly illustrated in @Michael Hardy's answer. So to sum up, you must make a factor $(x-4)$ in numerator, in order to cancel the same at denominator. But, if that factor was not there, how it comes after multiplication by conjugate ? The actual answer is We don't need $(x-4)$, but we need to cancel $(\sqrt{x}-2)$ (note that the other factor $(\sqrt{x}+2)$ gives no trouble) and surely, the numerator should have such a factor, though not explicitly. So, here is an alternate way (without multiplying by conjugate and this approach hopefully applicable to similar problems):
First set $y=4+3x$. Then as $x\to4,y\to16$ and yor limit becomes$$\lim_{y\to16}\frac{3}{y-16}.2\left(\frac{4}{\sqrt{y}}-1\right)$$ $$=\lim_{y\to16}\frac{6}{(\sqrt{y}-4)(\sqrt{y}+4)}\left(\frac{4-\sqrt{y}}{\sqrt{y}}\right)$$ $$=-\dfrac{3}{16}$$
