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I have about 10 of these problems I could post, can't figure any of them out...not sure what to do.

$\lim_{h \to 0}\dfrac{(\sqrt{9+h} -3)}{h}$

I know that I need to rationalize it so I multiply by the numerator which gives me

$9 + h - 9$ (not sure if positive or negative since it is $-3$ squared it should be $9$ but it could also be minus square of $3$ which would be -9 /h sqr(9+h) -3

From that I can divide by h and get 1/3-3

which I know is wrong. What am I doing wrong?

  • 0
    I was just going through my thought process.2011-08-27

5 Answers 5

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Posting as I havent seen anyone suggest Taylor expansion yet.

$\lim_{h \to 0}\dfrac{\sqrt{9+h} -3}{h} = \lim_{h \to 0}\dfrac{3\sqrt{1+h/9} -3}{h}$

Taylor expanding the numerator becomes $3(1+ \frac{1}{2}\cdot\frac{h}{9}) -3 = \frac{h}{6} + O(h^2)$

$\lim_{h\to 0} \dfrac{h/6+O(h^2)}{h} = \frac{1}{6}$

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    If the OP has not yet learnt about derivatives... (:2011-09-06
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If $\rm\ f(x)\: = \ f_0 + f_1\ x +\:\cdots\:+f_n\ x^n\:$ and $\rm\: f_0 \ne 0\:$ then rationalizing the numerator below yields

$\rm \lim_{x\:\to\: 0}\ \dfrac{\sqrt{f(x)}-\sqrt{f_0}}{x}\ = \ \lim_{x\:\to\: 0}\ \dfrac{f(x)-f_0}{x\ (\sqrt{f(x)}+\sqrt{f_0})}\ =\ \dfrac{f_1}{2\ \sqrt{f_0}}$

Your problem is the special case $\rm\ f(x) = 9 + x\ $ with $\rm\ f_0 =9,\ f_1 = 1\:,\:$ so the limit equals $\:1/6\:.\:$

When you study derivatives you'll see how they mechanize this process in a very general way. Namely the above limit is \rm\:g'(0)\ for $\rm\:g(x) = \sqrt{f(x)}\:,\:$ so applying general rules for calculating derivatives we easily mechanically calculate that \rm\:g'(x)\: =\: f\:\:'(x)/(2\:\sqrt{f(x)})\:.\: Evaluating it at $\rm\:x=0\:$ we conclude that \rm\: g'(0)\: =\: f\:\:'(0)/(2\:\sqrt{f(0)})\: =\: f_1/(2\:\sqrt{f_0})\:,\: exactly as above.

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In response to a comment by Jordan Carlyon. We start by rationalizing the numerator. To do that we multiply and divide $\sqrt{9+h}-3$ by $\sqrt{9+h}+3$. In general $\sqrt{a}-b$ would be rationalized by multiplying and dividing by $\sqrt{a}+b$.

$\begin{eqnarray*} \lim_{h\rightarrow 0}\frac{\sqrt{9+h}-3}{h} &=&\lim_{h\rightarrow 0}\frac{ \left( \sqrt{9+h}-3\right) \left( \sqrt{9+h}+3\right) }{h\left( \sqrt{9+h} +3\right) } \\ &=&\lim_{h\rightarrow 0}\frac{\left( \sqrt{9+h}\right) ^{2}-3^{2}}{h\left( \sqrt{9+h}+3\right) } \\ &=&\lim_{h\rightarrow 0}\frac{9+h-9}{h\left( \sqrt{9+h}+3\right) } \\ &=&\lim_{h\rightarrow 0}\frac{h}{h\left( \sqrt{9+h}+3\right) } \\ &=&\lim_{h\rightarrow 0}\frac{1}{\sqrt{9+h}+3} \\ &=&\frac{1}{\displaystyle\lim_{h\rightarrow 0}\sqrt{9+h}+3} \\ &=&\frac{1}{\sqrt{9}+3} \\ &=&\frac{1}{6} \end{eqnarray*}$

Alternatively, you may use use L'Hôpital's rule, as I wrote in my reply to your 1st question.

Added: Since you have "not learned about derivatives yet" and this rule uses the evaluation of derivatives, it is to be learned later.

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    @Bill: Thanks! The '%'s are a by-product of the way I generate the TeX code. I use Scientific Word and then I copy the code from a .tex file, using the Windows note book. I have to pay attention to that and remove those '%'s in the future.2011-08-28
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Consider $f:[0,\infty[\to \mathbb{R}:x\mapsto \sqrt{x}$. Then \begin{align*} f'(9)&=\lim_{h\to 0} \frac{f(9+h)-f(9)}{h}\\ &= \lim_{h\to 0} \frac{\sqrt{9+h}-3}{h},\\ f'(x)&= \frac{1}{2\sqrt{x}}, \end{align*} so f'(9)=\frac{1}{6}.

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    I have not learned about derivatives yet.2011-08-28
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Let $t=\sqrt{9+h}$. Then $h=t^2-9$ and $h \to 0$ iff $t \to 3$, so your limit is equal to $\lim_{t \to 3} \frac{t-3}{t^2 - 9}$. Since $t^2 -9 = (t-3)(t+3)$, this is just $\lim_{t \to 3} \frac{1}{t+3} = \frac{1}{6}$.