$3x+\frac{1}{3}>x+\frac{4}{5}$
$\lim\:_{x\to\:\:0}\left(\frac{x}{\left(\ln\:^3\left(x\right)+2\right)}\right)$
$tan\left(\frac{1}{4}x\right)=1$
$191.6-123.99$
$\left(2x^3-\frac{1}{5}\right)\left(2x^3+3\right)$
$\left(-3x^4y^2\right)^2$
$\lim_{x\to1}\left(\frac{x^3-6x+5}{4x^2+3x-4}\right)$
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