$2\cdot\lim_{t\to\infty}\left(t\cdot\ln\left(1+\frac{3}{t}\right)\right)$
$\lim_{x\to\infty}\left(\frac{sin\left(3x\right)}{x}\right)$
$8x+1=-23$
$\frac{x^2+2x-24}{x^2-9}\cdot\frac{x+3}{x^2-16}$
$4x-8<x-2$
$\left(2u^5v^2\right)^{-2}$
$\left(2j^3.4k^3\right)^2$
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