$$\lim_{x\to3}\left(\frac{x^2+4x}{x-3}\right)$$

$$12+\left(-8+7\right)-\left(4-7\right)$$

$$\frac{dy}{dx}+100y=100cos60x$$

$$5x^2-500$$

$$4g+6g-3g$$

$$\int\left(8x^2-8\right)^{\left(\frac{3}{2}\right)}dx$$

$$3y'+4y=32$$