A calculus problem by Abdulrahman El Shafei

$\begin{aligned} L & = \lim_{x \to \infty} x \sin \left( \frac 1x \right) & \small \color{#3D99F6} \text{Let } u = \frac 1x \\ & = \lim_{u \to 0} \frac {\sin u}u & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \lim_{u \to 0} \frac {\cos u}1 & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. } u \\ & = \boxed{1} \end{aligned}$

Let $x=99999$ , we have

$99999 \cdot \sin \left(\dfrac{1}{99999}\right) = 1$