$\lim_{x\to2}\left(\frac{2x-3}{x^2-4}\right)$
$\sqrt[2]{1.7}$
$\frac{6.4}{2.6}$
$\frac{1}{x^2+2x-24}$
$3\left(-14\right)+\left(-4\right)\left(-3\right)-4\left(6\right)$
$\lim_{x\to0\:}\left(\frac{tan\left(9x\right)}{tan\left(4x\right)}\right)$
$x-\frac{3}{4}\ge\left(x+1\right)\cdot\left(-4\right)$
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