$\frac{1}{\frac{x-1}{x+1}}$
$1+tan\left(x\right)=\frac{sec^2\left(x\right)+2tan\left(x\right)}{1+tan\left(x\right)}$
$-4x^4y^2.4x^{-2}y^{-3}$
$-7<10-26$
$\lim_{x\to2}\left(3x-7\right)$
$\lim_{x\to0}\left(\frac{\sqrt{x^2+9}}{x}\right)$
$\lim_{x\to0}\left(\frac{8\tan^2\left(\frac{1}{2}x\right)}{\frac{3}{2}x\sin\left(4x\right)}\right)$
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