$\lim_{x\to\infty}\left(1+\frac{0.1}{x}\right)^x$
$u^2-6u+9=0$
$x^2-128x+64$
$\left(u+\frac{3}{4}y\right)^{2}$
$0=6x-\frac{48}{x^2}$
$\cot4x^5\cdot\left(4x^2+3\right)$
$5x^2y^3z^4;\:x=-1;\:y=-1;\:z=-1$
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