1.6 Answer

1.6

1.

(a) $${\lim_{n \to \infty} n^4 - 3n^3 = \lim_{n \to \infty}n^{4}(1 - \frac{3}{n}) = \infty}$$

(b) $${\lim_{n \to \infty}\frac{3n^2 + 5}{4n^3 - 1} = \lim_{n \to \infty}\frac{n^2(3 + \frac{5}{n^2})}{n^3(4 - \frac{1}{n^3})} = 0}$$

(c) $${\lim_{n \to \infty}\frac{1 - n}{n - \sqrt{n}} = \lim_{n \to \infty}\frac{n(\frac{1}{n} - 1)}{n(1 - \frac{\sqrt{n}}{n})} = -1}$$

(d)

$$\lim_{n \to \infty}\frac{n(n+2)}{n+1} - \frac{n^3}{n^2 + 1} = \lim_{n \to \infty}\frac{(n^2 + 2n)(n^2 + 1) - n^3(n+1)}{(n+1)(n^2 + 1)}$$

$$= \lim_{n \to \infty}\frac{n^3 + n^2 + 2n}{n^3 + n^2 + n+ 1}$$

$$= \lim_{n \to \infty}\frac{n^3(1 + \frac{1}{n} + \frac{2}{n^2})}{n^3(1 + \frac{1}{n} + \frac{1}{n^2} + \frac{1}{n^3})} = 1$$

(e)

$$\lim_{n \to \infty}\sqrt{n+1} - \sqrt{n} = \lim_{n \to \infty}\frac{n+1 - n}{(\sqrt{n+1} + \sqrt{n})} = 0$$

2.

Note that for $a > 1$, we have $\sqrt[n]{a} > 1$. Thus, we can set $\sqrt[n]{a} = 1 + h, \ h > 0$. Then

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

Thus,

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

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

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

For $0 < a < 1$, we 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) $0 < b < a$ implies $a^{n} < a^{n} + b^{n} < 2a^{n}$． Then, $${a < (a^{n} + b^{n})^{\frac{1}{n}} < 2^{\frac{1}{n}} a}$$． Thus, $${\lim_{n \to \infty}(a^{n} + b^{n})^{\frac{1}{n}} = a}$$.

(b) Since $${3^{n} \leq 1 + 2^{n} + 3^{n} \leq 3 \dot 3^{n}}$$, we take $n$th root. Then $${3 \leq (1 + 2^{n} + 3^{n})^{\frac{1}{n}} \leq 3^{\frac{1}{n}} \dot 3}$$．As $n \to \infty$, $3^{\frac{1}{n}} \to 1$. Then by squeezing theorem, we have $${\lim_{n \to \infty}(1 + 2^{n} + 3^{n})^{\frac{1}{n}} = 3}$$