$\frac{tan\left(x\right)}{sec^2\left(x\right)}$

$x^2+1<2x$

$\lim_{x\to0}\left(\frac{cos\left(x+h\right)-cos\left(x\right)}{h}\right)$

$\log\left(x+6\right)-\log\left(2\right)=\log\left(x\right)$

$\int\left(16x+8\right)\left(4x^2+4x\right)dx$

$3.\:\log\left(x\right)=3$

$\lim_{x\to-2}\left(\frac{x^2+3x+2}{in\:\left(x+3\right)}\right)$