$\lim_{x\to\infty}\left(xe^{5x}-x\right)$
$6a+18b$
$n^2+16n+4$
$\left(\frac{2^3\cdot x}{a^5}\right)$
$x=y^6$
$-2:2+-9\cdot\left(-3\right)$
$l\lim_{x\to0}\:\frac{sin^2\left(2x\right)}{x^2}$
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