Use the squeeze theorem to evaluate the limit as x approaches 0 of x^2 sin(1/x).

Use the squeeze theorem to evaluate the limit as x approaches 0 of x^2 sin(1/x).

• A. 0
• B. DNE
• C. 1
• D. -1

Answer: (A)

The key idea is that sin(1/x) stays between -1 and 1 no matter how close x gets to zero, so the oscillation is bounded. This lets the squeeze theorem do the heavy lifting: for all x ≠ 0, -x^2 ≤ x^2 sin(1/x) ≤ x^2 because |sin(1/x)| ≤ 1. As x approaches 0, both outer expressions -x^2 and x^2 converge to 0. Since the expression in the middle is trapped between two quantities that both go to 0, it must also go to 0. Equivalently, |x^2 sin(1/x)| ≤ x^2 and x^2 → 0, so the limit is 0.

The key idea is that sin(1/x) stays between -1 and 1 no matter how close x gets to zero, so the oscillation is bounded. This lets the squeeze theorem do the heavy lifting: for all x ≠ 0, -x^2 ≤ x^2 sin(1/x) ≤ x^2 because |sin(1/x)| ≤ 1. As x approaches 0, both outer expressions -x^2 and x^2 converge to 0. Since the expression in the middle is trapped between two quantities that both go to 0, it must also go to 0. Equivalently, |x^2 sin(1/x)| ≤ x^2 and x^2 → 0, so the limit is 0.