$-16-+91$
$\lim_{x\to\infty}\:\frac{4^x}{x^3}$
$\left(a^2b^3\right)\cdot\left(-a^3b^4c\right)$
$\lim_{x\to0.01}\frac{1-\cos\left(0.01\right)}{0.01}$
$x+2<-4x+5$
$\left(-3x+3y\right)^4$
$\tan\left(x\right)+2\sin^2\left(x\right)$
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