1. **Problem Statement:** Find the Limit Inferior (\(\liminf\)) and Limit Superior (\(\limsup\)) of the sequences: \(a)\ (z_n) = (-2)^n\) \(b)\ (w_n) = 1 + \frac{(-1)^n}{n}\) \(c)\ (u_n) = \sin\left(\frac{n\pi}{3}\right)\) \(d)\ (v_n) = (1, 2, 3, 4, 1, 2, 3, 4, \ldots)\) 2. **Key Definitions:** - The Limit Superior \(\limsup_{n \to \infty} a_n\) is the supremum (least upper bound) of the set of subsequential limits. - The Limit Inferior \(\liminf_{n \to \infty} a_n\) is the infimum (greatest lower bound) of the set of subsequential limits. 3. **Sequence (a):** \(z_n = (-2)^n\) - The terms alternate between positive and negative powers of 2: \(2, -4, 8, -16, 32, -64, \ldots\) - The subsequence of even \(n\) is \(2^{2k} = 4^k \to +\infty\). - The subsequence of odd \(n\) is \(-2^{2k+1} = -2 \cdot 4^k \to -\infty\). - Therefore, \(\limsup z_n = +\infty\) and \(\liminf z_n = -\infty\). 4. **Sequence (b):** \(w_n = 1 + \frac{(-1)^n}{n}\) - As \(n \to \infty\), \(\frac{1}{n} \to 0\). - For even \(n\), \((-1)^n = 1\), so \(w_{2k} = 1 + \frac{1}{2k} \to 1^+\). - For odd \(n\), \((-1)^n = -1\), so \(w_{2k+1} = 1 - \frac{1}{2k+1} \to 1^-\). - The subsequential limits are \(1\) from above and below. - Hence, \(\limsup w_n = 1\) and \(\liminf w_n = 1\). 5. **Sequence (c):** \(u_n = \sin\left(\frac{n\pi}{3}\right)\) - The values repeat every 6 terms because \(\sin\left(\frac{(n+6)\pi}{3}\right) = \sin\left(\frac{n\pi}{3} + 2\pi\right) = \sin\left(\frac{n\pi}{3}\right)\). - The 6 values are: \(\sin(0) = 0\), \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\), \(\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}\), \(\sin(\pi) = 0\), \(\sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}\), \(\sin\left(\frac{5\pi}{3}\right) = -\frac{\sqrt{3}}{2}\). - The subsequential limits are these six values. - The maximum is \(\frac{\sqrt{3}}{2}\), the minimum is \(-\frac{\sqrt{3}}{2}\). - So, \(\limsup u_n = \frac{\sqrt{3}}{2}\) and \(\liminf u_n = -\frac{\sqrt{3}}{2}\). 6. **Sequence (d):** \(v_n = (1, 2, 3, 4, 1, 2, 3, 4, \ldots)\) - This sequence is periodic with period 4. - The subsequential limits are the values \(1, 2, 3, 4\). - The maximum is 4, the minimum is 1. - Hence, \(\limsup v_n = 4\) and \(\liminf v_n = 1\). **Final answers:** - \(\limsup z_n = +\infty, \liminf z_n = -\infty\) - \(\limsup w_n = 1, \liminf w_n = 1\) - \(\limsup u_n = \frac{\sqrt{3}}{2}, \liminf u_n = -\frac{\sqrt{3}}{2}\) - \(\limsup v_n = 4, \liminf v_n = 1\)