Using the squeeze theorem, what is lim x->0 of x^2 cos(1/x)?

Using the squeeze theorem, what is lim x->0 of x^2 cos(1/x)?

• A. 0
• B. DNE
• C. Zero
• D. Infinity

Answer: (A)

The key idea is the squeeze theorem. Since cos(1/x) always lies between -1 and 1, we have -1 ≤ cos(1/x) ≤ 1. Multiplying by x^2 gives -x^2 ≤ x^2 cos(1/x) ≤ x^2. As x → 0, both -x^2 and x^2 approach 0, so the middle expression must also approach 0 by squeezing. Hence the limit is 0. This matches the notion of zero, and the other possibilities (DNE or infinity) don’t fit because the product is bounded by numbers that go to zero.

The key idea is the squeeze theorem. Since cos(1/x) always lies between -1 and 1, we have -1 ≤ cos(1/x) ≤ 1. Multiplying by x^2 gives -x^2 ≤ x^2 cos(1/x) ≤ x^2. As x → 0, both -x^2 and x^2 approach 0, so the middle expression must also approach 0 by squeezing. Hence the limit is 0. This matches the notion of zero, and the other possibilities (DNE or infinity) don’t fit because the product is bounded by numbers that go to zero.