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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.

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    I assume that you are being asked to find the slope of the tangent line using only the *definition* of the derivative. If not, there is a far easier way. After a certain number of basic differentiation formulas and procedures have been established, we can use these procedures and formulas. Then the process of finding the slope of the tangent line involves far less ingenuity, and you can read off the answer from the formula for the derivative.2011-11-07
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    The expression $\lim\limits_{x \to 4} \dfrac{f(x)-f(4)}{x-4}$ will _always_ be of the form $0/0$. As long as you've still got $0/0$, you're not done. You have the fraction $1/(x-4)$, so you want to factor the thing your multiplying it by as $(x-4)\cdot(\text{something})$ and then cancel.2011-11-08
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    @Srivatsan: It looks from the problem, that the numerator $8$ should not be within the surds, in title.2011-11-08
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    Thanks @Swapan for the (good!) catch. Apologies to Pedro for the typo.2011-11-08

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