Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 8 Exercise 8.7

Solution:

### Part a

Suppose the limit exists and equals $a$. Then for $\epsilon=\dfrac{1}{2}$, there exists $N>0$ such that

$$

\left|\cos\left(\frac{n\pi}{3}\right)-a\right|<\frac{1}{2}

$$ for all $n>N$. Let $k>N$ be an integer, then we also have by taking $n=6k\pi$ and $n=6k\pi+3\pi$ respectively

$$

\left|\cos\left(\frac{6k\pi}{3}\right)-a\right|<\frac{1}{2},\quad \left|\cos\left(\frac{6k\pi+3\pi}{3}\right)-a\right|<\frac{1}{2}.

$$ Hence we have

$$

|1-a|<\frac{1}{2},\quad |-1-a|<\frac{1}{2}.

$$ Thus by Triangle Inequality we have

$$

2=|1-a-(-1-a)|<|1-a|+|-1-a|<1

$$ which is impossible. We obtain a contradiction. Thus the limit does not exist.

### Part b

Suppose the limit exists and equals $a$. Then for $\epsilon=\dfrac{1}{2}$, there exists $N>0$ such that

$$

\left|(-1)^nn-a\right|<\frac{1}{2}

$$ for all $n>N$. Let $k>N$ be an integer, then we also have by taking $n=2k$ and $n=2k+1$ respectively

$$

\left|(-1)^{2k}2k-a\right|<\frac{1}{2},\quad \left|(-1)^{2k+1}(2k+1)-a\right|<\frac{1}{2}.

$$ Hence we have

$$

|2k-a|<\frac{1}{2},\quad |-2k-1-a|<\frac{1}{2}.

$$ Thus by Triangle Inequality we have

$$

4k+1=|2k-a-(-2k-1-a)|<|2k-a|+|-2k-1-a|<1

$$ which is impossible. We obtain a contradiction. Thus the limit does not exist.

### Part c

Suppose the limit exists and equals $a$. Then for $\epsilon=\dfrac{1}{2}$, there exists $N>0$ such that

$$

\left|\sin\left(\frac{n\pi}{3}\right)-a\right|<\frac{1}{2}

$$ for all $n>N$. Let $k>N$ be an integer, then we also have by taking $n=6k\pi$ and $n=6k\pi+3\pi/2$ respectively

$$

\left|\sin\left(\frac{6k\pi}{3}\right)-a\right|<\frac{1}{2},\quad \left|\sin\left(\frac{6k\pi+3\pi/2}{3}\right)-a\right|<\frac{1}{2}.

$$ Hence we have

$$

|0-a|<\frac{1}{2},\quad |1-a|<\frac{1}{2}.

$$ Thus by Triangle Inequality we have

$$

1=|0-a-(1-a)|<|0-a|+|1-a|<1

$$ which is impossible. We obtain a contradiction. Thus the limit does not exist.