by using squeezing theorem we can prove that lim (x-->1) x sin (1/x) = 0?

Answer 1

No, you cannot, because [x→1]lim x sin (1/x) = 1*sin(1/1) = sin 1 ≠ 0. You may, however, use the squeeze theorem to prove that [x→0]lim x sin (1/x) = 0, by simply noting that for any x≠0, -|x| ≤ x sin (1/x) ≤ |x|, and that [x→0]lim -|x| = [x→0]lim |x| = 0.