tính các giới hạn sau

Question

a)

$\mathop {\lim }\limits\frac{2\cos n^{2}}{n^{2}+1}$
b)

$\mathop {\lim }\limits\frac{3\sin n-4\cos n}{2n^{2}+1}$
c)

$\mathop {\lim }\limits\frac{(-1)^{n}\sin (3n+n^{2})}{3n-1}$
d)

$\mathop {\lim }\limits\frac{2-2n\cos n}{3n+1}$

score 1 · Answer 1

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

$\lim -\dfrac{5}{2n^2+1} =\lim \dfrac{5}{2n^2+1} = 0 \Rightarrow \lim \dfrac{3\sin x -4\cos x}{2n^2+1}=0$

1 Đáp án