RHL=$\lim_{x\rightarrow 0^{+}} sin(1/x)$ put $x\rightarrow 0+h$ then, RHL= $\lim_{h\rightarrow 0} sin(1/h)$
it gives positive value {-1 to 1}
LHL = $\lim_{x\rightarrow 0^{-}} sin(1/x)$ put $x\rightarrow 0-h$ then ,LHL=$\lim_{h\rightarrow 0} sin(-1/h) = \lim_{h\rightarrow 0} -sin(1/h)$ it gives negative value [-1 to 1]
so LHL !=RHL so limit doesnot exist