$\lim_{x\to\infty}\left(\frac{ln\left(x+4\right)}{x+2}\right)$
$\left(-3x^{-2}y^4\right)^3$
$\left(3x^n\:+\:2\right)^3\:\:$
$-x^2-x+3<0$
$6x^3-12x^2+6x$
$3x+2x-x+2^2$
$4+\left(4-\left(5-3\right)\right)$
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