$\lim_{x\to\infty}\left(\frac{ln\left(9x+2\right)}{ln\left(5x+4\right)+6}\right)$
$cos^2=cos2$
$x^2-4x+-4$
$\lim_{x\to-\infty}\left(\frac{\left(x^2+6x+9\right)}{x^2+9}\right)$
$x^3-x^2-5x+6$
$4x^2\:+3x^2-4x$
$x^2+7x-18=0$
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