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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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    Thanks @Swapan for the (good!) catch. Apologies to Pedro for the typo.2011-11-08

5 Answers 5

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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 ?

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    @Pedro $a^{-1} = \frac{1}{a}$. I was simply typesetting trick to avoid multi-story fractions.2011-11-08
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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.

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    Thanks.My issue was that I was expected a final expression more like to the derivative formula.I was trying to force the expression to be like the derivative formula, but it seems they are not equal.2011-11-07
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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 $x$.

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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:

$\lim_{x \to 4}\frac{f(x)-f(4)}{x-4}=\lim_{x \to 4}-\frac{12}{(3 x+4)^{3/2}}$

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    @ArturoMagidin I think it's right what you are saying. But I think you agree that it's a valid process too and maybe he didn't need find the answer using the definition. You choose @Pedro^^2011-11-09
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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$\lim_{y\to16}\frac{3}{y-16}.2\left(\frac{4}{\sqrt{y}}-1\right)$ $=\lim_{y\to16}\frac{6}{(\sqrt{y}-4)(\sqrt{y}+4)}\left(\frac{4-\sqrt{y}}{\sqrt{y}}\right)$ $=-\dfrac{3}{16}$

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    Please let me know if you feel this to be *simple* method2011-11-08