Evaluate the limit as x approaches 0 of x sin(1/x) using the squeeze theorem.

Evaluate the limit as x approaches 0 of x sin(1/x) using the squeeze theorem.

• A. DNE
• B. 0
• C. 1
• D. -1

Answer: (B)

The key idea is that sin(1/x) stays between -1 and 1 no matter how x behaves near 0. Multiply by x, and you get -|x| ≤ x sin(1/x) ≤ |x|. As x approaches 0, both -|x| and |x| go to 0, so the whole expression is squeezed to 0. This shows the limit exists and equals 0, because the oscillations of sin(1/x) are damped by the factor x shrinking to zero. The magnitude cannot approach 1 or -1, and it certainly doesn’t diverge, since it’s always within ±|x|, which collapses to 0.

The key idea is that sin(1/x) stays between -1 and 1 no matter how x behaves near 0. Multiply by x, and you get -|x| ≤ x sin(1/x) ≤ |x|. As x approaches 0, both -|x| and |x| go to 0, so the whole expression is squeezed to 0. This shows the limit exists and equals 0, because the oscillations of sin(1/x) are damped by the factor x shrinking to zero. The magnitude cannot approach 1 or -1, and it certainly doesn’t diverge, since it’s always within ±|x|, which collapses to 0.