$\lim_{x\to5}\frac{x^2-5x-24}{x^2-3x-18}$
$\lim_{x\to\infty}\left(\frac{16x^2-6x-5}{8x^2-5x+4}\right)$
$2 + 5 - ( 1 - 5 ) - 2 + ( 2 + 3 )$
$\frac{\sin\theta}{1+\sin\theta}-\frac{\sin\theta}{1-\sin\theta}$
$4x^2+3x-2xy$
$-16sin4x=0$
$\frac{3x}{x-1}>2$
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