a) Ta có $-\dfrac{2}{n^2 +1} \le \dfrac{2\cos n^2}{n^2 +1} \le \dfrac{2}{n^2+1}$
Mà $\lim (-\dfrac{2}{n^2 +1})=\lim (\dfrac{2}{n^2 +1})=0 \Rightarrow \lim ( \dfrac{2\cos n^2}{n^2 +1}) =0 $
b) Ta có $-\sqrt{3^2 +4^2} =-5 \le 3\sin x -4\cos x \le \sqrt{32+4^2}=5$
Vậy $- \dfrac{5}{2n^2+1} \le \dfrac{3\sin x -4\cos x}{2n^2+1} \le \dfrac{5}{2n^2+1}$
mà $\lim -\dfrac{5}{2n^2+1} =\lim \dfrac{5}{2n^2+1} = 0 \Rightarrow \lim \dfrac{3\sin x -4\cos x}{2n^2+1}=0$