$4t\left(1-t\right)^3$
$\lim_{x\to\infty}\left(\frac{7-6x^3}{-7-6x^2}\right)$
$\lim_{x\to4}\left(\frac{x^2-3x+4}{x^2-16}\right)$
$-18x+8=-x^2$
$7-x+3x-3-8x$
$\left(2.5x^2+8y^2\right)^2$
$\lim_{x\to\infty}\left(\sqrt{x^2+x+1}-\sqrt{x^2+2x+2}\right)$
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