Let $f\left(x\right)$ be a bounded function. $L_1=\underset{x\to \infty }{\lim }\left(f^{\text{'}}\left(x\right)-\lambda f\left(x\right)\right)$ and $L_2=\underset{x\to \infty }{\lim }f\left(x\right)$ where $\mathrm{\lambda }>0.$ If $L_1, L_2$ both exist and $L_1=L,$ then prove that $L_2=-\frac{L}{\lambda }$

Consider the following statements: 

${\mathbf{S}}_1:$  $\underset{x\to 0^{-}}{\lim }\frac{\left[x\right]}{x}$ is an indeterminate form (where $[.]$ denotes greatest integer function).

${\mathbf{S}}_2:$  $\underset{x\to \infty }{\lim }\frac{\sin \left(3^x\right)}{3^x}=0$

${\mathbf{S}}_3:$  $\underset{x\to \infty }{\lim }\sqrt{\frac{x-\sin x}{x+{\cos }^{2}x}}$ does not exist.

${\mathbf{S}}_4:$  $\underset{n\to \infty }{\lim }\frac{(n+2)!+(n+1)!}{(n+3)!}\left(n\in N\right)=0$

State, in order, whether ${\mathrm{S}}_1,{\mathrm{S}}_2,{\mathrm{S}}_3,{\mathrm{S}}_4$ are true or false