$\lim_{x\to\infty}\left(\frac{5x+1}{e^x}\right)$
$x^5+x^3\cdot y^2=4$
$\left(9x^2-6\right).\left(9x^2+6\right)$
$-3=9$
$\lim_{h\to0}\left(\frac{10}{x+1}\right)$
$\int\frac{x^2+6x+1}{2x\left(x^2+1\right)}dx$
$x-24\ge9$
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