$\lim_{x\to\infty}\left(\frac{\sin6x}{\sin5x}\right)$
$11+5+6+4-7-6+4$
$2\tan\left(x\right)+1=0$
$\lim_{x\to0}\left(\frac{sin\left(x\right)}{x^2}\right)^{\frac{1}{x}}$
$-\frac{3}{2}m^2n;\:m=2;\:n=-7$
$\left(-6+3\right)\left(-5+4-9\right)$
$x-7\le8$
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