$f\left(x\right)=03+1$
$\lim_{x\to\infty}\left(\frac{x^3+2x}{e^x}\right)$
$x+\frac{1}{x\:}=3$
$\left(gx-6\right)\left(gx+6\right)$
$\lim_{x\to\infty}-\left(x+1\right)e^{-x}$
$\lim_{x\to0}\left(\frac{e^x-x}{2e^x-x^2-2}\right)$
$6n^4\:-\:12n^2\:+\:6$
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