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## Show the limit does not exist

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.