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So I feel stupid for asking this, but I can't figure this out. I haven't taken algebra for about 8 years, so doing this is kind of fuzzy.

Just started Calc 1 and we're finding limits. $$\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} .$$

I try to do some algebra to rationalize the denominator, but everything I do gets me to the limit equaling either $2$ or $3$. Which makes me think I don't understant rationalizing the denominator.

What I get is: $$\lim_{x \to 9} \frac{x\sqrt{x} - 9\sqrt{x}}{x - 3\sqrt{x}}$$

This is where I'm confusing myself. I don't know where to go to simplify from here. And I still can't do direct substitution because it will equal $\frac{0}{0}$.

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marked as duplicate by Normal Human, Arctic Char, Joel Reyes Noche, S.Panja-1729, Alex M. Sep 10 '15 at 7:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

The back of my book says the answer is 6... but I have no idea where that is coming from. – OghmaOsiris Jun 9 '11 at 19:02
Multiply the expression by $(\sqrt{x}+3)/(\sqrt{x}+3)$ to rationalize the denominator. You want to multiply both the numerator and denominator by the conjugate of the denominator to rationalize the denominator. – yunone Jun 9 '11 at 19:03
Ok, I was just multiplying by $\sqrt{x}$. I see where I went wrong. Thanks! – OghmaOsiris Jun 9 '11 at 19:06
The thing to remember is that $(a+b)(a-b)=a^2-b^2$. Similarly with complex denominators, remember that $(a+ib)(a-ib)=a^2+b^2$. – Henry Jun 9 '11 at 19:48
up vote 3 down vote accepted

To rationalize the denominator $\sqrt{x}-3$, you should multiply by $\sqrt{x}+3$. That way you get $$(\sqrt{x}-3)(\sqrt{x}+3) = \sqrt{x}\sqrt{x} - 3\sqrt{x} + 3\sqrt{x}-9 = x-9.$$ So we have: $$\lim_{x\to 9}\frac{x-9}{\sqrt{x}-3} = \lim_{x\to 9}\frac{(x-9)(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)} = \lim_{x\to 9}\frac{(x-9)(\sqrt{x}+3)}{x-9}.$$ Now, although this evaluates to $\frac{0}{0}$, note that because $x$ is approaching $9$ but not equal to $9$, then $x-9$ is not actually zero, so you can cancel the $x-9$ factor in the numerator with the one in the denominator.

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HINT $\ $ For $\rm\ z = \sqrt{x}\ $ the fraction is $\rm\displaystyle\ \frac{z^2-9}{z-3}\ = \ \cdots$

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This makes no sense to me. – OghmaOsiris Jun 9 '11 at 19:11
@ogh If you can elaborate why it makes no sense to you I'll be happy to explain further. What is not clear? – Bill Dubuque Jun 9 '11 at 19:14
I don't know what you're hinting at or how this relates to the problem. All I see is you making z = $\sqrt{x}$ – OghmaOsiris Jun 9 '11 at 19:15
@ogh Notice that the denominator divides the numerator (either using difference of squares or the Factor Theorem). So simply perform the division, then undo the variable change, and you have your rationalized result (more simply than brute-force application of the rule of multiplying the numerator and denominator by the conjugate). When such exact divisions occur they will generally be more efficient than the conjugate method. – Bill Dubuque Jun 9 '11 at 19:20
@Ogh Yes, I'm happy to see it is clear now. My hints are often on the terse side since I think that leads to better opportunities for learning experiences (vs. serving a complete answer on a silver platter). – Bill Dubuque Jun 9 '11 at 19:54

Hint: Multiply the top at the bottom by $\sqrt{x}+3$. This will "rationalize" the denominator since $(\sqrt{x}+3)(\sqrt{x}-3)=x-9$.

Hope that helps,

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