$12-2\left(4\right)\left(3\right)+\left(9\right)$
$\lim_{x\to0}\left(1+\frac{1}{3x}\right)$
$-2\cdot3+4\cdot5$
$2x^2-3x=0$
$\lim\:_{x\to\:\:0}\:\frac{9x^2}{ln\left(sec\:x\right)}$
$\left(3x+1\right)\left(9x^2-3x+1\right)$
$sin=\frac{2}{3}$
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