$\lim_{x\to\infty}\left(\frac{x-1}{x^2+6x+4}\right)$
$\lim_{x\to\infty}\left(\frac{\ln\left(x\right)}{x^2+3}\right)$
$\left(-58\right)\:-\:\left(-63\right)$
$\sqrt[4]{64y^4}z^{13}$
$\left(x-5\right)\left(x-7\right)\left(x+6\right)\left(x+4\right)-504$
$14-\left(-18\right)$
$70>y+40$
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