1.5 Sequences

1.

(a) $\infty$(b) 0(c) $-1$ (d) 1(e) 0

2.

For $a > 1$, $\sqrt[n]{a} > 1$. Thus let $\sqrt[n]{a} = 1 + h, h > 0$. Then

$$a = (1 + h)^{n} = 1 + nh + \cdots + h^n \geq 1 + nh$$

Therefore

$$\lim_{n \rightarrow \infty}h = \lim_{n \rightarrow \infty}\frac{a - 1}{n} = 0$$

For $a = 1$, $\sqrt[n]{a} = 1$ for all $n$. Thus

$$\lim_{n \rightarrow \infty}\sqrt[n]{a} = 1$$

For $0 < a < 1$, let $${b = \frac{1}{a}}$$. Then $b > 1$ and

$$1 = \lim_{n \rightarrow \infty}\sqrt[n]{b} = \lim_{n \rightarrow \infty}\frac{1}{\sqrt[n]{a}}$$

3.

(a) a(b) 3