$\frac{\sqrt{n}\left(2x+1\right)^{3n}}{n^2}$
$x2\:+\:5x\:+\:6\:\le0\:$
$-\left(2-7\right)$
$\frac{k}{k+1}+\frac{1}{\left(k+1\right)\left(k+2\right)}=\frac{k+1}{k+2}$
$\lim_{x\to\infty\:}\left(2-x\right)$
$\frac{x^2-xy}{x}$
$\frac{x^4}{4+4x^2}$
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