Question

Prove that `lim_(x→0)xsin(2/x)=0` ?

Answer 1

`-1 <= sin(2/x) <= 1` `" "` for all `x != 0`

On the right

For `x > 0` we multiply the inequality by `x` to get

`-x < xsin(2/x) < x` `" "` for all `x != 0`

`lim_(xrarr0^+) -x = 0` `" "` and `" "` `lim_(xrarr0^+) x = 0`.

Therefore, by the Squeeze Theorem (right limit) we have

`lim_(xrarr0^+)xsin(2/x) = 0` `" "` (Eq 1)

On the left

For `x < 0` we multiply the inequality by `x` and we must reverse the inequalities to get

`-x > xsin(2/x) > x` `" "` for all `x != 0`

`lim_(xrarr0^-) -x = 0` `" "` and `" "` `lim_(xrarr0^-) x = 0`.

Therefore, by the Squeeze Theorem (left limit) we have

`lim_(xrarr0^-)xsin(2/x) = 0` `" "` (Eq 2)

The two-sided limit,

By Eq 1 and 2, we get

`lim_(xrarr0)xsin(2/x) = 0`