a. $limx_{n}^{'}=lim\frac{1}{2n\pi }=0$
$limx_{n}^{''}=lim\frac{1}{\left ( 2n+1 \right )\frac{\pi }{2}}=0$
$limf\left ( x_{n}^{'} \right )=limcos\frac{1}{\frac{1}{2n\pi }}=limcos2n\pi =1$
$limf\left ( x_{n}^{''} \right )=lim cos\frac{1}{\frac{1}{\left ( 2n+1 \right )\frac{\pi }{2}}}= lim cos\left ( 2n+1 \right )\frac{\pi }{2}=0$
b. Ta có :
$x_{n}^{'}=\frac{1}{2n\pi }\rightarrow 0, x_{n}^{''}=\frac{1}{\left ( 2n+1 \right )\frac{\pi}{2}\rightarrow 0}$
nhưng $f\left ( x_{n}^{'} \right )\rightarrow 1\neq 0\leftarrow f\left ( x_{n}^{''} \right )$
Do đó không tồn tại