$\lim_{x\to\infty}\left(\frac{x+1}{x^3+x}\right)$
$\int_{-1}^1\left(\frac{4x+2}{\left(x-2\right)\left(x+3\right)}\right)dx$
$\frac{dy}{dx}+1=8x-y$
$3x^2-4x-6;\:x=-2$
$4x+6\ge2x-3$
$\lim_{x\to-3}\left(\frac{x^2+6x+9}{-24-5x-x^2}\right)$
$0.8\cdot39+4.2$
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