$\lim_{x\to\infty}\left(\frac{x+4}{x}\right)^x$
$f\left(x\right)=ln\sqrt{1-x^8}$
$-5.93-8.62-\left(-5.93\right)$
$\sqrt{1^2}2^2$
$-521\:.\:\left(+15\right)$
$\left(2x+y\right)^{2}=2x+y^{2}$
$\lim_{x\to2}\left(\frac{x-2}{2\sqrt{2+x}}\right)$
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