Make substitution `t=1/x` so if `x->0,` then `t->oo` .

`lim_(x->0)x sin(1/x)=lim_(t->oo)1/tsin(t)`

Sine is bounded function (`sin(t) in [-1,1]`) and `lim_(t->oo)1/t=0`

therefore `lim_(t->oo)1/tsin(t)=0.` Hence

`lim_(x->0)x sin(1/x)=0` **<-- Your solution**

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