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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 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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    @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.