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.
 A: 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 ?
A: Just note that 
$$\begin{align}
\frac{8}{\sqrt{4 + 3x}} - 2 &= \frac{8 - 2 \sqrt{4 + 3x}}{\sqrt{4 + 3x}}
\\
&= \frac{8 - 2 \sqrt{4 + 3x}}{\sqrt{4 + 3x}} \cdot \frac{8 + 2\sqrt{4 + 3x}}{8 + 2 \sqrt{4 + 3x}}
\\
&= \frac{64 - 4(4 + 3x)}{\sqrt{4 + 3x}(8 +  \sqrt{4 +3x})}.
\end{align}$$
I'll leave the rest up to you.
A: 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 $x$.
A: 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:
$$\lim_{x \to 4}\frac{f(x)-f(4)}{x-4}=\lim_{x \to 4}-\frac{12}{(3 x+4)^{3/2}}$$
A: 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$$\lim_{y\to16}\frac{3}{y-16}.2\left(\frac{4}{\sqrt{y}}-1\right)$$ $$=\lim_{y\to16}\frac{6}{(\sqrt{y}-4)(\sqrt{y}+4)}\left(\frac{4-\sqrt{y}}{\sqrt{y}}\right)$$ $$=-\dfrac{3}{16}$$
