$\lim_{x\to\infty}\frac{2x^3}{e^{4x}}$
$\left(5-mn^2\right)^3\:$
$-9-3c+5c-4$
$\left(\frac{2}{3x}\right).\frac{1}{4y}\cdot\frac{2}{1}$
$\lim_{x\to1}\left(\frac{x^2-7x+6}{x^2-5x+6}\right)$
$16x^22+20xy+25y^22$
$25x^2-4=0$
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