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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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  • $\begingroup$ 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. $\endgroup$ – André Nicolas Nov 7 '11 at 22:17
  • $\begingroup$ 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. $\endgroup$ – Michael Hardy Nov 8 '11 at 1:25
  • $\begingroup$ @Srivatsan: It looks from the problem, that the numerator $8$ should not be within the surds, in title. $\endgroup$ – Tapu Nov 8 '11 at 2:57
  • $\begingroup$ Thanks @Swapan for the (good!) catch. Apologies to Pedro for the typo. $\endgroup$ – Srivatsan Nov 8 '11 at 2:59
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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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  • $\begingroup$ Our answers are much in the same (there's only so many ways to do things), but you were quicker than me. Let me know if you want me to delete my answer. $\endgroup$ – JavaMan Nov 7 '11 at 22:10
  • $\begingroup$ @JavaMan No need to delete, the path to simplification is a little different. $\endgroup$ – Sasha Nov 7 '11 at 22:14
  • $\begingroup$ Your way to put the conjugate is curious.I took a few minutes to figure it out what was that negative exponent $\endgroup$ – Pedro Nov 7 '11 at 23:16
  • $\begingroup$ @Pedro $a^{-1} = \frac{1}{a}$. I was simply typesetting trick to avoid multi-story fractions. $\endgroup$ – Sasha Nov 8 '11 at 0:07
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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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  • $\begingroup$ 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. $\endgroup$ – Pedro Nov 7 '11 at 23:09
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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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    $\begingroup$ Given that the poster is being asked for the derivative of $f(x)$ at the point $4$ (via the definition), I would think that invoking L'Hopital's Rule is certainly not okay, and essentially circular: if you know how to compute $f'(x)$, then why are you computing $f'(4)$ using a limit? This is like trying to justify $\lim_{x\to 0}\frac{\sin x}{x}=1$ by using L'Hopital's Rule, when you need to know this limit in order to know what $(\sin x)'$ is in the first place. $\endgroup$ – Arturo Magidin Nov 8 '11 at 2:19
  • $\begingroup$ @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^^ $\endgroup$ – GarouDan Nov 9 '11 at 21:11
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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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  • $\begingroup$ Please let me know if you feel this to be simple method $\endgroup$ – Tapu Nov 8 '11 at 2:50

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