$\lim_{x\to\infty}\left(\frac{3x^4+6x^3}{2x^3+3x}\right)$
$6\:\left(x\:+\:1\right)\:+\:\left(-5\right)\:.\:\left(-4\right)\:=\:3\:\left(2x\:+\:4\right)\:-\:2\:.\:\left(-10\right)$
$\frac{5+3x}{4x+5}>1$
$ax^2+6axy+9ay^2$
$\lim_{x\to-1}\left(x-5\right)$
$2\left(-2\right)+6$
$16x^2-36y^2$
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