<s>[INST] Approach to calculating limit via polynomial division Given the following exercise:
$ \lim_{x \to \frac{1}{3}} (\frac{27x^2-6x-1}{6x^2+x-1}) $
The rule of L'Hospital with the first derivative yields:
$\lim_{x \to \frac{1}{3}} \frac{27 x^2 - 6 x - 1}{6 x^2 + x - 1} = \frac{12}{5}$
Another approach would be factoring out the x which also yields:
$\frac{12}{5}$
However, I'm asking myself whether polynomial division, too, could help solving the limit.
Polynomial division yields:
$27 x^2 - 6 x - 1 = \frac{9}{2} \cdot (6 x^2 + x - 1) + \frac{7}{2} - \frac{21x}{2}$
polynomial division on Wolfram Alpha
Thanks.
 [/INST] Hint:  To use polynomial division, I would first write (for $x \neq \frac13$):
$$\frac{27x^2-6x-1}{6x^2+x-1} = \frac92 - \frac7{2(2x+1)}$$
Now taking the limit is trivial...
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