$\lim_{x\to0}\frac{1-\cos\left(4x\right)}{x^2}$

$100-\left(-16\right)$

$\left(3xy+y^2\right)+\left(x^2+xy\right)\frac{dy}{dx}=0$

$3x-2>x+5$

$c^2-5c+12$

$f\left(x\right)=\frac{1}{\left(x^6-6x\right)^3}$

$\frac{3\left(\infty\right)^2+2\infty-15}{-16\left(\infty\right)}$