$$\lim_{x\to\infty}\left(\frac{\ln\left(4\right)}{1+\ln\left(x\right)}\right)$$

$$x^2-4>6$$

$$9x^2-4x+4$$

$$y\left[n+2\right]-4y\left[n+1\right]+3y\left[n\right]$$

$$\frac{d}{dx}xy+x=2y$$

$$\frac{d}{dx}\left(5x^4-x^2y^3+7y^5=-10\right)$$