Convergence of $\lbrace \sin( \frac{1}{n}) \rbrace $ I believe that the sequence $\lbrace  \sin( \frac{1}{n}) \rbrace $ converges to zero. Can someone give an episilon-delta proof of this fact?
 A: We have
$$ \forall x \in \mathbb{R} : x \geq 0 : \sin(x) \leq x$$
Note that $\forall n \in \mathbb{N} :    \pi > 1 \geq \frac{1}{n} \geq 0$. Hence combining the two results we have
$$ \forall n \in \mathbb{N} :  0 < \sin\left(\frac{1}{n}\right) \leq \frac{1}{n} \implies \left| \sin\left(\frac{1}{n}\right)\right| \leq \frac{1}{n}$$
Now for the epsilon-delta. Let $\varepsilon > 0$ be arbitrary. Choose $N > \frac{1}{\varepsilon}$
$$ \forall n \geq N : \left| \sin\left(\frac{1}{n}\right)\right| \leq \frac{1}{n} \leq \frac{1}{N}  < \frac{1}{\frac{1}{\varepsilon}} = \varepsilon$$
Note that the first inequality is regarded as the "small-angle approximation". It can be derived geometrically, or by using the mean value theorem on the function defined by $f(x) = \sin(x) - x$.
After reading the comments on the question I also like the method used by the user hcl14! Note that the derivative of $\sin(x)$ is bounded by $1$ and hence the sine function is Lipschitz continuous with Lipschitz constant $L = 1$. This implies that 
$$\left|\sin\left(\frac{1}{x}\right) - \sin(0)\right| \leq \left|\frac{1}{x} - 0\right| \iff \left|\sin\left(\frac{1}{x}\right)\right| < \frac{1}{x}$$
A: By the mean value theorem,
$$\left|\sin\left(\frac1n\right)-\sin0\right|=|\cos\xi|\cdot\left|\frac1n-0\right|\leq\frac1n$$
for some $\xi\in[0,\frac1n]$. So for $n>\delta:=\frac1\epsilon$ we have $|\sin\frac1n|<\epsilon$.
