<s>[INST] 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.
 [/INST] 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$$\lim_{y\to16}\frac{3}{y-16}.2\left(\frac{4}{\sqrt{y}}-1\right)$$ $$=\lim_{y\to16}\frac{6}{(\sqrt{y}-4)(\sqrt{y}+4)}\left(\frac{4-\sqrt{y}}{\sqrt{y}}\right)$$ $$=-\dfrac{3}{16}$$
</s>