$\lim_{x\to0}\left(\cot\left(8x\right)-\frac{1}{8x}\right)$
$\left(2x^2-3x^3+4x^4\right)^2$
$\left(-51\right)\cdot\left(-21\right)$
$8\cdot13mn^2$
$\frac{2x^2}{x^2+4}$
$\lim_{x\to\infty}\left(\frac{ln10^x}{4^x}\right)$
$90+8^{0.5}$
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