$\lim_{x\to\infty}\left(\frac{x-1}{x-4}\right)^x$
$5x-8=2x+4$
$\frac{5secx}{secx-tanx}=5secx\left(secx+tanx\right)$
$\frac{d}{dx}\left(\frac{8x^2}{2x^4+5}\right)$
$2x^2+5x+1=0$
$-4+27+16+36$
$\frac{3x^5+2x^2-5}{x^5-3x^3+8}$
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