Hadi B. answered • 06/20/19

BS in Mathematics with 5+ years of tutoring experience

since sine is always between -1 and 1, consider the following inequality:

lim_{(x→0)} x(-1) ≤ lim_{(x→0)} x sin 1/x ≤ lim_{(x→0)} x(1)

lim_{(x→0)} -x ≤ lim_{(x→0)} x sin 1/x ≤ lim_{(x→0)} x

use the squeeze theorem to show the limit in the middle is 0 too