Which one of the following function defined below is continuous at origin?

(A) $\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\sin \frac{1}{x}=$   does not exist, Because $\sin \frac{1}{x}$  oscillates from $-1$  to 1 in the neighborhood of $x=0.$   So $f\left(x\right)$  is discontinuous at   $x=0.$
(B) $\underset{x\to 0}{\lim }g\left(x\right)=\frac{1}{6}\cos 6=g\left(0\right)\Rightarrow g\left(x\right)$  is continuous at  $x=0.$
(C)  $\underset{x\to 0}{\lim }h\left(x\right)=\underset{x\to 0}{\lim }x.\left({\tan }^{-1}\frac{1}{x}\right)=0.\left(\frac{-\pi }{2}to\frac{\pi }{2}\right)=0$
But, $h\left(0\right)=1$ , $h\left(x\right)$  is discontinuous at  $x=0.$
(D)  $\underset{x\to 0}{\lim }k\left(x\right)=\sin 1$, But  $k\left(0\right)=1$
So,  $k\left(x\right)$ is discontinuous at  $x=0.$