# Finding $\lim\limits_{h \to 0}\frac{(\sqrt{9+h} -3)}{h}$

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?

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I have formatted part of your question - but I cannot parse your fourth line (not sure if positive or negative...). What did you mean there? –  mixedmath Aug 27 '11 at 23:51
I was just going through my thought process. –  user138246 Aug 27 '11 at 23:54

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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According to wikipedia that rule states that the limit of a quotient is just the limit of the numerator divided by the limit of the denominator correct? And if so how do I find their limits? What happened in the last part of that problem? All of a sudden it jumps from the square root into 6, where does six come from and why is it +3 and not -3? –  user138246 Aug 28 '11 at 0:03
@Jordan Carlyon: It is when both limits are different from zero. –  Américo Tavares Aug 28 '11 at 0:05
... which does not happen here. You have two options, manipulate the quotient, or use L'Hôpital's rule. –  Américo Tavares Aug 28 '11 at 0:06
For a little while, there will be questions from the very beginnings of a calculus course. It should not be difficult to recognize such questions. In particular, something that comes in an introduction to derivatives should not be attacked using L'Hospital's Rule! Here unfortunately the book begins with a special section on limits, instead of melding the appropriate limit calculations seamlessly into a calculation of the derivative. "Rationalizing the numerator" is the intended approach. –  André Nicolas Aug 28 '11 at 1:13
@Américo The '%'s in TeX in your answer and comments are causing it not to render in IE. I "fixed" this in you answer, but I'm not sure what you intend. What's the '%' for? –  Bill Dubuque Aug 28 '11 at 18:48

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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+1 for your generalization. –  Américo Tavares Aug 28 '11 at 1:15

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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@Jordan Carlyon: this is useful to you if you know the definition of derivative. –  leo Aug 28 '11 at 0:32
I have not learned about derivatives yet. –  user138246 Aug 28 '11 at 0:38

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 $\lim_{t \to 3} \frac{1}{t+3} = \frac{1}{6}$.

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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... (: –  JavaMan Sep 6 '11 at 22:46