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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:


integer over variable results in 0:

12/ (0 - 8) =

12/-8 =


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? – 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
up vote 2 down vote accepted

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.

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