$\lim_{h\to0}\left(\frac{\sqrt[3]{x+1}-1}{x}\right)$
$\lim_{x\to\infty}\left(\sqrt{\frac{n^2-n+2}{6n-8}}\right)$
$v^3\:=\:-1$
$\sqrt{16n^2}$
$5+11-5-2-9+8+6+8-4-3-8+5+6+5-7$
$\int\frac{1}{x^2\sqrt{x^2+a}}dx$
$x^3+xy$
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