Giải thích các bước giải:
\(\begin{array}{l}
15,\\
\lim \left( {\sqrt {{n^2} - 4n + 7} + a - n} \right)\\
= \lim \left[ {\left( {\sqrt {{n^2} - 4n + 7} - n} \right) + a} \right]\\
= \lim \left[ {\frac{{{n^2} - 4n + 7 - {n^2}}}{{\sqrt {{n^2} - 4n + 7} + n}} + a} \right]\\
= \lim \left[ {\frac{{ - 4n + 7}}{{\sqrt {{n^2} - 4n + 7} + n}} + a} \right]\\
= \lim \left[ {\frac{{ - 4 + \frac{7}{n}}}{{\sqrt {1 - \frac{4}{n} + \frac{7}{{{n^2}}}} + 1}} + a} \right]\\
= \frac{{ - 4}}{{\sqrt 1 + 1}} + a\\
= a - 2\\
\lim \left( {\sqrt {{n^2} - 4n + 7} + a - n} \right) = 0 \Leftrightarrow a - 2 = 0 \Leftrightarrow a = 2\\
16,\\
\lim \left( {\sqrt[3]{{{n^3} + 9{n^2}}} - n} \right)\\
= \lim \left( {\frac{{{n^3} + 9{n^2} - {n^3}}}{{{{\sqrt[3]{{{n^3} + 9{n^2}}}}^2} + \sqrt[3]{{{n^3} + 9{n^2}}}.n + {n^2}}}} \right)\\
= \lim \frac{{9{n^2}}}{{{{\sqrt[3]{{{n^3} + 9{n^2}}}}^2} + n.\sqrt[3]{{{n^3} + 9{n^2}}} + {n^2}}}\\
= \lim \frac{9}{{{{\sqrt[3]{{1 + \frac{9}{n}}}}^2} + 1.\sqrt[3]{{1 + \frac{9}{n}}} + 1}}\\
= \frac{9}{{1 + 1 + 1}} = 3\\
17,\\
\lim \left( {\sqrt[3]{{{n^3} + {n^2} - 1}} - 3\sqrt {4{n^2} + n + 1} + 5n} \right)\\
= \lim \left[ {\left( {\sqrt[3]{{{n^3} + {n^2} - 1}} - n} \right) + 3.\left( {2n - \sqrt {4{n^2} + n + 1} } \right)} \right]\\
= \lim \left[ {\frac{{{n^3} + {n^2} - 1 - {n^3}}}{{{{\sqrt[3]{{{n^3} + {n^2} - 1}}}^2} + n.\sqrt[3]{{{n^3} + {n^2} - 1}} + {n^2}}} + 3.\frac{{4{n^2} - 4{n^2} - n - 1}}{{2n + \sqrt {4{n^2} + n + 1} }}} \right]\\
= \lim \left[ {\frac{{{n^2} - 1}}{{{{\sqrt[3]{{{n^3} + {n^2} - 1}}}^2} + n.\sqrt[3]{{{n^3} + {n^2} - 1}} + {n^2}}} + 3.\frac{{ - n - 1}}{{2n + \sqrt {4{n^2} + n + 1} }}} \right]\\
= \lim \left[ {\frac{{1 - \frac{1}{{{n^2}}}}}{{{{\sqrt[3]{{1 + \frac{1}{n} - \frac{1}{{{n^3}}}}}}^2} + 1.\sqrt[3]{{1 + \frac{1}{n} - \frac{1}{{{n^3}}}}} + 1}} + 3.\frac{{ - 1 - \frac{1}{n}}}{{2 + \sqrt {4 + \frac{1}{n} + \frac{1}{{{n^2}}}} }}} \right]\\
= \frac{1}{3} + \frac{{ - 3}}{4} = \frac{{ - 5}}{{12}}
\end{array}\)
