Using the Squeeze Theorem, evaluate lim_{x->0} x^2 cos(1/x).

Using the Squeeze Theorem, evaluate lim_{x->0} x^2 cos(1/x).

• A. 0
• B. 1
• C. -1
• D. Does not exist

Answer: (A)

The key idea is to use the Squeeze Theorem. Because cos(1/x) is always between -1 and 1, multiplying by x^2 gives -x^2 ≤ x^2 cos(1/x) ≤ x^2 for all x ≠ 0. As x approaches 0, both -x^2 and x^2 approach 0. Since the function in the middle is trapped between two expressions that both go to 0, its limit must be 0. Therefore the limit is 0. It can’t be 1 or -1, and it does exist, because the squeeze forces it to 0.

The key idea is to use the Squeeze Theorem. Because cos(1/x) is always between -1 and 1, multiplying by x^2 gives -x^2 ≤ x^2 cos(1/x) ≤ x^2 for all x ≠ 0. As x approaches 0, both -x^2 and x^2 approach 0. Since the function in the middle is trapped between two expressions that both go to 0, its limit must be 0. Therefore the limit is 0. It can’t be 1 or -1, and it does exist, because the squeeze forces it to 0.