$x-1\le3x-5$
$\lim_{x\to\infty}\left(\frac{x^2+1}{x+1}\right)$
$2y^5z^2\cdot y^4z^5$
$-\left(-\frac{1}{4\left(3\infty^2+2\right)^2}\right)$
$sin\theta\:+\frac{cos\theta\:}{tan\theta\:}=\frac{1}{cos\theta\:\cdot\:\:tan\theta\:}$
$+2-1-6-4$
$\frac{3}{y}=\frac{6}{y-2}$
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