Limit

U Sub L'Hopital

\begin{gathered} \lim_{ x \to \infty } x \sin \frac{1}{x} \\ u = \frac{1}{x} \\ \lim_{ x \to \infty } x \frac{u}{u} \sin u \\ \lim_{ x \to \infty } x u \cdot \lim_{ x \to \infty } \frac{\sin u}{u} \\ \lim_{ x \to \infty } x \frac{1}{x} \cdot 1 \\ 1 \end{gathered}

\begin{gathered} \frac{d}{dx} \frac{\tan x}{x} \\ \frac{\sec^2 x}{x} - \frac{\tan x}{x^2} \end{gathered}

\begin{gathered} \frac{d}{dx} x \sin \left( \frac{1}{x} \right) \\ \sin \left( \frac{1}{x} \right) - x\frac{ \cos \left( \frac{1}{x} \right) }{x^2} \\ \sin\left( \frac{1}{x} \right)-\frac{\cos\left( \frac{1}{x} \right)}{x} \end{gathered}

\begin{align} \frac{dy}{dx}(x^2y^3+\sin y &= 4) \\ 2xy^3 + 3x^2 y ^2 \frac{dy}{dx} + \frac{dy}{dx}\cos y &= 0 \\ \frac{dy}{dx}(3x^2y^2 + \cos y) &= -2xy^3 \\ \frac{dy}{dx} = -\frac{2xy^3}{3x^2y^2 + \cos y} \end{align}

\begin{gathered} \int (x^2 + 2x) \cos( x^3 + 3x^2 - 47 ) \ dx \\ u = x^3 + 3x^2 - 47 \\ \frac{du}{dx} = 3x^2 + 6x \\ \frac{1}{3} \int \frac{du}{dx} \cos u \ dx \\ \frac{1}{3} \int \cos u \ du \\ \frac{1}{3} \sin u + C \\ \frac{1}{3} \sin (x^3 + 3x^2 - 47) + C \end{gathered}