$\lim \left( {\sqrt[3]{{{n^3} - 3{n^2} + 1}} - \sqrt {{n^2} + 4n} } \right)=-3$

$\lim \dfrac{{\sin 10n + \cos 10n}}{{{n^2} + n}} = 0$

Giải thích các bước giải:

\(\begin{array}{l}
a,\\
\lim \left( {\sqrt[3]{{{n^3} - 3{n^2} + 1}} - \sqrt {{n^2} + 4n} } \right)\\
 = \lim \left[ {\left( {\sqrt[3]{{{n^3} - 3{n^2} + 1}} - n} \right) + \left( {n - \sqrt {{n^2} + 4n} } \right)} \right]\\
 = \lim \left( {\frac{{{n^3} - 3{n^2} + 1 - {n^3}}}{{{{\sqrt[3]{{{n^3} - 3{n^2} + 1}}}^2} + \sqrt[3]{{{n^3} - 3{n^2} + 1}}.n + {n^2}}} + \frac{{{n^2} - {n^2} - 4n}}{{n + \sqrt {{n^2} + 4n} }}} \right)\\
 = \lim \left( {\frac{{ - 3{n^2} + 1}}{{{{\sqrt[3]{{{n^3} - 3{n^2} + 1}}}^2} + \sqrt[3]{{{n^3} - 3{n^2} + 1}}.n + {n^2}}} + \frac{{ - 4n}}{{n + \sqrt {{n^2} + 4n} }}} \right)\\
 = \lim \left( {\frac{{ - 3 + \frac{1}{{{n^2}}}}}{{{{\sqrt[3]{{1 - \frac{3}{n} + \frac{1}{{{n^2}}}}}}^2} + \sqrt[3]{{1 - \frac{3}{n} + \frac{1}{{{n^2}}}}} + 1}} + \frac{{ - 4}}{{1 + \sqrt {1 + \frac{4}{n}} }}} \right)\\
 = \frac{{ - 3}}{{1 + 1 + 1}} + \frac{{ - 4}}{{1 + 1}} =  - 1 - 2 =  - 3\\
\end{array}\)

b) Ta có

$\sin(10n) + \cos(10n) = \sqrt{2} \sin\left( 10n + \dfrac{\pi}{4} \right)$

Suy ra

$\left\vert \sqrt{2} \sin\left( 10n + \dfrac{\pi}{4} \right) \right\vert \leq \sqrt{2}$

Suy ra

$0 \leq \lim \dfrac{\left\vert \sqrt{2} \sin\left( 10n + \frac{\pi}{4} \right) \right\vert}{n^2 +n} \leq \lim \dfrac{\sqrt{2}}{n^2 + n} = 0$

Vậy 

$ \lim \dfrac{\left\vert \sqrt{2} \sin\left( 10n + \frac{\pi}{4} \right) \right\vert}{n^2 +n} = 0$

Suy ra

$\lim \dfrac{\sin (10n) + \cos(10n)}{n^2 + n} = 0$