tính giới hạn a) lim √(9n ^2+3n-4) -3n+2 b) lim ( ∛(n^3+3n^2) -n c) cho a ∈ R và lim( √(an^2+n+4) -2n+1=+ ∞

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
\lim \left( {\sqrt {9{n^2} + 3n – 4}  – 3n + 2} \right)\\
 = \lim \left( {\frac{{9{n^2} + 3n – 4 – {{\left( {3n – 2} \right)}^2}}}{{\sqrt {9{n^2} + 3n – 4}  + 3n – 2}}} \right)\\
 = \lim \frac{{9{n^2} + 3n – 4 – 9{n^2} + 12n – 4}}{{\sqrt {9{n^2} + 3n – 4}  + 3n – 2}}\\
 = \lim \frac{{15n – 8}}{{\sqrt {9{n^2} + 3n – 4}  + 3n – 2}}\\
 = \lim \frac{{15 – \frac{8}{n}}}{{\sqrt {9 + \frac{3}{n} – \frac{4}{{{n^2}}}}  + 3 – \frac{2}{n}}}\\
 = \frac{{15}}{{\sqrt 9  + 3}} = \frac{{15}}{6} = \frac{5}{2}\\
b,\\
\lim \left( {\sqrt[3]{{{n^3} + 3{n^2}}} – n} \right)\\
 = \lim \left( {\frac{{{{\sqrt[3]{{{n^3} + 3{n^2}}}}^3} – {n^3}}}{{{{\sqrt[3]{{{n^3} + 3{n^2}}}}^2} + n.\sqrt[3]{{{n^3} + 3{n^2}}} + {n^2}}}} \right)\\
 = \lim \frac{{3{n^2}}}{{{{\sqrt[3]{{{n^3} + 3{n^2}}}}^2} + n.\sqrt[3]{{{n^3} + 3{n^2}}} + {n^2}}}\\
 = \lim \frac{3}{{\sqrt[3]{{1 + \frac{3}{n}}} + 1.\sqrt[3]{{1 + \frac{3}{n}}} + 1}} = \frac{3}{{1 + 1 + 1}} = 1
\end{array}\)