Bài tập 13 trang 142 SGK Toán 11 NC

Bài tập 13 trang 142 SGK Toán 11 NC

Ta có:

\(\begin{array}{*{20}{l}}
{2n + \cos n = n\left( {2 + \frac{{\cos n}}{n}} \right)}\\
\begin{array}{l}
\left| {\frac{{\cos n}}{n}} \right| \le \frac{1}{n},\lim \frac{1}{n} = 0\\
 \Rightarrow \lim \frac{{\cos n}}{n} = 0
\end{array}
\end{array}\)

Do đó \(\lim \left( {2 + \frac{{\cos n}}{n}} \right) = 2 > 0\) và \(\lim n =  + \infty \)

Suy ra \(\lim \left( {2n + \cos n} \right) =  + \infty \)

b)

\(\begin{array}{l}
\lim \left( {\frac{1}{2}{n^2} - 3\sin 2n + 5} \right)\\
 = \lim {n^2}\left( {\frac{1}{2} - \frac{{3\sin 2n}}{{{n^2}}} + \frac{5}{{{n^2}}}} \right) =  + \infty 
\end{array}\)

(vì \(\lim {n^2} =  + \infty ,\lim \left( {\frac{1}{2} - \frac{{3\sin 2n}}{{{n^2}}} + \frac{5}{{{n^2}}}} \right) = \frac{1}{2} > 0\))

-- Mod Toán 11