$\lim_{x\to0}\left(\frac{8^x-1}{sin2x}\right)$
$\lim_{x\to0}\left(\frac{e^x}{xe^x}\right)$
$x^2+8x=8$
$\frac{d}{dx}\sqrt{x^2+4}$
$\lim\:_{x\to\:\:1}\left(\frac{x^2-1}{\sqrt{x+3}-2}\right)$
$16^2-40t+25$
$-1\int\sqrt{1-\sin\left(2x\right)}dx$
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