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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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    The back of my book says the answer is 6... but I have no idea where that is coming from.2011-06-09
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    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.2011-06-09
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    Ok, I was just multiplying by $\sqrt{x}$. I see where I went wrong. Thanks!2011-06-09
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    The thing to remember is that $(a+b)(a-b)=a^2-b^2$. Similarly with complex denominators, remember that $(a+ib)(a-ib)=a^2+b^2$.2011-06-09

3 Answers 3

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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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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.2011-06-09
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    @ogh If you can elaborate why it makes no sense to you I'll be happy to explain further. What is not clear?2011-06-09
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    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}$2011-06-09
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    @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.2011-06-09
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    This works for this problem, but if the denominator doesn't divide the numerator perfectly, I can't use this short cut. Its a good shortcut, tho. I'll keep it in mind. Thanks2011-06-09
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    @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).2011-06-09
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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.