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Calculate $lim_{x \to (\frac{9}{8})} \frac{12x}{9-8x}$ and $lim_{x\to(-\frac{9}{8})} \frac{12x}{9-8x}$.

I am confused how to do this. L'hospital, substitution and conjugates will not solve this as they result in the division of zero. I tried factoring out the largest degree:

(x/x)(12/[9/x]-8)

integer over variable results in 0:

12/ (0 - 8) =

12/-8 =

-3/2

However, upon further investigation this is the limit as it approaches infinity. I have tried a few other things, but none have resulted in a correct answer, this is as close as I have gotten. What do I have to do to get $x \to \frac{9}{8}$ instead of $x \to (+)\infty$? Thanks.

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    Using wolfram|alpha I obtained the correct solutions: (-)infinity, (+)infnity; respectively for +9/8 and -9/8. However, I still do not understand how to obtain this solution algebraically?2011-02-16
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    Just look at what happens when $x \to \frac{9}{8}+$. The denominator $\to 0$ and hence it goes to $-\infty$.2011-02-16
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    Yes, I understand now...thanks both of you! :)2011-02-16

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If the limit of the denominator is 0 and the limit of the numerator is nonzero, the limit does not exist (is $\pm\infty$). You can easily determine the sign by plugging in numbers close to $\frac{9}{8}$ on either side.