$x\:\frac{dy}{dx}+4y\:=\:x^3\:-\:x$

$\lim_{x\to0}\left(\frac{1-\cos\left(2x\right)sin5x}{x^2sin3x}\right)$

$-3x\left(4+2x\right)$

$3\left(2+6p\right)$

$\lim_{x\to0}\left(\frac{t-sin\left(t\right)}{1-cos\left(t\right)}\right)$

$\left(4ab^2+xy\right)^2$

$\frac{1}{16x^3+4x^2}$