1.$\underset{x->0}{\lim }sin(\frac{1}{x})$的极限不存在

取数列f(x),$x_n=\frac{1}{2n\Pi -\frac{\Pi }{2}}$和$y_n=\frac{1}{2n\Pi +\frac{\Pi }{2}}$

$\underset{x->\infty }{\lim }sin(x_n)=\underset{x->0}{\lim }sin(\frac{1}{2n\Pi -\frac{\Pi }{2}})=-1$
$\underset{x->\infty }{\lim }sin(y_n)=\underset{x->0}{\lim }sin(\frac{1}{2n\Pi +\frac{\Pi }{2}})=1$
$$\begin{array}{l}\underset{x->\infty }{\lim }{x}_n=0,x_n\ne 0\\ \underset{y->\infty }{\lim }{y}_n=0,y_n\ne 0\\ \underset{x->\infty }{\lim }sin(x_n)\ne \underset{x->\infty }{\lim }sin(y_n)\end{array}\}=>\underset{x->0}{\lim }sin(\frac{1}{x})极限不存在$$

2.$\underset{x->0}{\lim }sin(x)$的极限不存在

$\underset{x->\infty }{\lim }sin(x_n)=\underset{x->0}{\lim }sin(2n\Pi -\frac{\Pi }{2})=-1$
$\underset{x->\infty }{\lim }sin(y_n)=\underset{x->0}{\lim }sin(2n\Pi +\frac{\Pi }{2})=1$

$$\begin{array}{l}\underset{x->\infty }{\lim }{x}_n=0,x_n\ne 0\\ \underset{y->\infty }{\lim }{y}_n=0,y_n\ne 0\\ \underset{x->\infty }{\lim }sin(x_n)\ne \underset{x->\infty }{\lim }sin(y_n)\end{array}\}=>\underset{x->\ast }{\lim }sin(x)极限不存在$$