$\lim_{x\to\infty}\:\left(2-x\right)\left(3+7x\right)$

$\frac{\sin^2\left(x\right)}{\tan^2\left(x\right)}=1-\sin^2\left(x\right)$

$\left(5n^2+6p^3\right)\cdot\left(5n^2-6p^3\right)$

$\frac{x^4-x}{3x^2}$

$\frac{d^3y}{dx}\:y=6x^2-4x+2\:$

$20z^2+4=0$

$\lim_{x\to\infty}\left(\frac{\left(6x^2-9x+8\right)}{\left(3x^2+2\right)}\right)$