$f\left(x\right)=\left(\sqrt{x^2+1}\right)^3$
$5\sin^2-1=\sin^2$
$16x^{-0.5}$
$\lim_{x\to2}\left(\frac{x^3-8}{x^2-x-2}\right)$
$\lim\:_{x\to\:1}\frac{\left(\sqrt{2x+1}-3\right)}{x-1}$
$\frac{1.43\cdot10^3}{8.3\cdot10^5}$
$-2\cdot2-3$
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