$\frac{x^{2}+4x+4}{x^{2}-1}:\frac{x+2}{x+1}$
$7+\left(16-4^2\right)-8$
$\frac{x}{2}+\frac{2}{3}=\frac{7}{6}$
$\lim_{x\to0}\left(\frac{sin\left(x^3\right)}{sin^3\left(x\right)}\right)$
$2\cdot\frac{x}{3}$
$\frac{x^4-x^3+x^2-3x-6}{x+3}$
$\lim_{x\to-\infty}\left(2x-1-\sqrt{4x^2+4x+4}\right)$
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