$\lim_{x\to\infty}\left(\frac{x^2lnx}{x^2+lnx}\right)$

$6\:x\frac{1}{3}$

$8x=4x^2-1$

$2x+6<0$

$ln\:\left(\frac{\left(3-x\right)}{\left(3+x\right)}\right)$

$\sec^4\left(x\right)-2\sec^2\left(x\right)\cdot\tan^2\left(x\right)+\tan^4\left(x\right)$