1. **Statement of the problem:** We have three sequences defined for $n \in \mathbb{N}^*$: $$U_n = \sqrt{4n^2 + 5}, \quad V_n = n^2 + 2n - 1, \quad W_n = n + 5.$$ We want to find the limits as $n \to +\infty$ of: $$U_n, \quad V_n, \quad W_n, \quad U_n - 2W_n, \quad U_n - V_n, \quad U_n - W_n.$$ 2. **Calculate $\lim_{n \to +\infty} U_n$:** Rewrite $U_n$: $$U_n = \sqrt{4n^2 + 5} = \sqrt{4n^2\left(1 + \frac{5}{4n^2}\right)} = 2n \sqrt{1 + \frac{5}{4n^2}}.$$ As $n \to +\infty$, $\frac{5}{4n^2} \to 0$, so: $$\sqrt{1 + \frac{5}{4n^2}} \to 1,$$ thus: $$\lim_{n \to +\infty} U_n = \lim_{n \to +\infty} 2n \times 1 = +\infty.$$ 3. **Calculate $\lim_{n \to +\infty} V_n$:** Since $V_n = n^2 + 2n - 1$, the dominant term is $n^2$, which grows without bound: $$\lim_{n \to +\infty} V_n = +\infty.$$ 4. **Calculate $\lim_{n \to +\infty} W_n$:** Since $W_n = n + 5$, which grows linearly: $$\lim_{n \to +\infty} W_n = +\infty.$$ 5. **Calculate $\lim_{n \to +\infty} (U_n - 2W_n)$:** Recall: $$U_n = 2n \sqrt{1 + \frac{5}{4n^2}} = 2n \left(1 + \frac{5}{8n^2} + o\left(\frac{1}{n^2}\right)\right) = 2n + \frac{5}{4n} + o\left(\frac{1}{n}\right).$$ Also: $$2W_n = 2(n + 5) = 2n + 10.$$ So: $$U_n - 2W_n = \left(2n + \frac{5}{4n} + o\left(\frac{1}{n}\right)\right) - (2n + 10) = \frac{5}{4n} - 10 + o\left(\frac{1}{n}\right).$$ As $n \to +\infty$, $\frac{5}{4n} \to 0$ and $o\left(\frac{1}{n}\right) \to 0$, so: $$\lim_{n \to +\infty} (U_n - 2W_n) = -10.$$ 6. **Calculate $\lim_{n \to +\infty} (U_n - V_n)$:** Recall: $$U_n = 2n + \frac{5}{4n} + o\left(\frac{1}{n}\right),$$ and $$V_n = n^2 + 2n - 1.$$ Then: $$U_n - V_n = \left(2n + \frac{5}{4n} + o\left(\frac{1}{n}\right)\right) - (n^2 + 2n - 1) = -n^2 + 1 + \frac{5}{4n} + o\left(\frac{1}{n}\right).$$ As $n \to +\infty$, $-n^2 \to -\infty$, so: $$\lim_{n \to +\infty} (U_n - V_n) = -\infty.$$ 7. **Calculate $\lim_{n \to +\infty} (U_n - W_n)$:** Recall: $$U_n = 2n + \frac{5}{4n} + o\left(\frac{1}{n}\right),$$ and $$W_n = n + 5.$$ Then: $$U_n - W_n = \left(2n + \frac{5}{4n} + o\left(\frac{1}{n}\right)\right) - (n + 5) = n - 5 + \frac{5}{4n} + o\left(\frac{1}{n}\right).$$ As $n \to +\infty$, $n - 5 \to +\infty$, so: $$\lim_{n \to +\infty} (U_n - W_n) = +\infty.$$ **Final answers:** $$\lim_{n \to +\infty} U_n = +\infty,$$ $$\lim_{n \to +\infty} V_n = +\infty,$$ $$\lim_{n \to +\infty} W_n = +\infty,$$ $$\lim_{n \to +\infty} (U_n - 2W_n) = -10,$$ $$\lim_{n \to +\infty} (U_n - V_n) = -\infty,$$ $$\lim_{n \to +\infty} (U_n - W_n) = +\infty.$$