Bài tập 18 trang 143 SGK Toán 11 NC

a)

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

b)

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

Câu c:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\lim \left( {\sqrt {{n^2} + n + 2}  - \sqrt {n + 1} } \right)\\
 = \lim n.\left( {\sqrt {1 + \frac{1}{n} + \frac{2}{{{n^2}}}}  - \sqrt {\frac{1}{n} + \frac{1}{{{n^2}}}} } \right) =  + \infty 
\end{array}
\end{array}\)

(vì \(\begin{array}{l}
\lim n =  + \infty ,\lim \left( {\sqrt {1 + \frac{1}{n} + \frac{2}{{{n^2}}}}  - \sqrt {\frac{1}{n} + \frac{1}{{{n^2}}}} } \right) = 1 > 0
\end{array}\))

d)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\lim \frac{1}{{\sqrt {3n + 2}  - \sqrt {2n + 1} }}\\
 = \lim \frac{{\sqrt {3n + 2}  + \sqrt {2n + 1} }}{{3n + 2 - 2n - 1}}
\end{array}\\
\begin{array}{l}
 = \lim \frac{{\sqrt {3n + 2}  + \sqrt {2n + 1} }}{{n + 1}}\\
 = \lim \frac{{n\left( {\sqrt {\frac{3}{n} + \frac{2}{{{n^2}}}}  + \sqrt {\frac{2}{n} + \frac{1}{{{n^2}}}} } \right)}}{{n\left( {1 + \frac{1}{n}} \right)}}
\end{array}\\
{ = \lim \frac{{\sqrt {\frac{3}{n} + \frac{2}{{{n^2}}}}  + \sqrt {\frac{2}{n} + \frac{1}{{{n^2}}}} }}{{1 + \frac{1}{n}}} = 0}
\end{array}\)

e)

\(\begin{array}{l}
\lim \left( {\sqrt {n + 1}  - \sqrt n } \right)n = \lim \sqrt n .\frac{{\sqrt n }}{{\sqrt {n + 1}  + \sqrt n }}\\
 = \lim \sqrt n .\frac{1}{{\sqrt {1 + \frac{1}{n}}  + 1}} =  + \infty 
\end{array}\)

(vì \(\lim \sqrt n  =  + \infty ,lim\frac{1}{{\sqrt {1 + \frac{1}{n}}  + 1}} = \frac{1}{2} > 0\))

f)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\lim \frac{{\sqrt {{n^2} + 1}  - \sqrt {n + 1} }}{{3n + 2}}\\
 = \lim \frac{{n\left( {\sqrt {1 + \frac{1}{{{n^2}}}}  - \sqrt {\frac{1}{n} + \frac{1}{{{n^2}}}} } \right)}}{{n\left( {3 + \frac{2}{n}} \right)}}
\end{array}\\
{ = \lim \frac{{\sqrt {1 + \frac{1}{{{n^2}}}}  - \sqrt {\frac{1}{n} + \frac{1}{{{n^2}}}} }}{{3 + \frac{2}{n}}} = \frac{1}{3}}
\end{array}\)

-- Mod Toán 11 HỌC247