Limits of positive and negatives?

limit: [(12x)/(9-8x)] as x --> (+)9/8 and (-)9/8. 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 --> (+)9/8 instead of x --> (+)inf? 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? –  Mr_CryptoPrime Feb 16 '11 at 9:22
Just look at what happens when $x \to \frac{9}{8}+$. The denominator $\to 0$ and hence it goes to $-\infty$. –  anonymous Feb 16 '11 at 9:26
Yes, I understand now...thanks both of you! :) –  Mr_CryptoPrime Feb 16 '11 at 9:29

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.