Transcrição 03/10/2025 14:18:15

Anotacoes

• Subtopic: Inequalities involving cosine function -1 < cos(1/x) ≤ 1 Multiplying all parts of the inequality by x² (which is greater than 0 for x ≠ 0): -1/(x²) < x²·cos(1/x) ≤ 1/x² This simplifies to: -x² < x²·cos(1/x) ≤ x² And further: -x² ≤ E(x) ≤ x² where E(x) = x²·cos(1/x). --- • Subtopic: Limits as x approaches 0 We observe that: lim (x→0) of x² = 0 and lim (x→0) of -x² = 0 By the Squeeze Theorem (T.C.), since: -x² ≤ E(x) ≤ x² and both bounds tend to 0 as x approaches 0, it follows that: lim (x→0) E(x) = 0 Therefore, the limit of x²·cos(1/x) as x approaches 0 is 0. --- • Subtopic: Final conclusion Hence, by the Squeeze Theorem, we conclude: lim (x→0) x²·cos(1/x) = 0