# Show that `lim_(x->0) xsin(1/x) = 0` I know what the graph looks like, but I have trouble proving it algebraically. Please help!

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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**

` `

`|sin(x)|<=1` for all `x in RR`

also

`|x|<=|x|` by property of real numbers.

Thus

`|x||sin(1/x)|<=|x| AA x in RR`

`|xsin(1/x)|<=|x|`

`lim_(x->0)|xsin(1/x)|<=lim_(x->0)|x|`

`lim_(x->0)|xsin(1/x)|<=0`

But positive number can not be less than zero.

Thus ,By definition of limit

`lim_(x->0)xsin(1/x)=0`