Tính các giới hạn sau :
a. \(\lim \sqrt {3{n^4} - 10n + 12} \)
b. \(\lim \left( {{{2.3}^n} - {{5.4}^n}} \right)\)
c. \(\lim \left( {\sqrt {{n^4} + {n^2} + 1} - {n^2}} \right)\)
d. \(\lim \frac{1}{{\sqrt {{n^2} + 2n} - n}}\)
a. \(\lim \sqrt {3{n^4} - 10n + 12} \)
\( = \lim {n^2}.\sqrt {3 - \frac{{10}}{{{n^3}}} + \frac{{12}}{{{n^4}}}} = + \infty \)
b. \(\lim \left( {{{2.3}^n} - {{5.4}^n}} \right) = \lim {4^n}\left[ {2{{\left( {\frac{3}{4}} \right)}^n} - 5} \right] = - \infty \)
c.
\(\begin{array}{l}
\lim \left( {\sqrt {{n^4} + {n^2} + 1} - {n^2}} \right)\\
= \lim \frac{{{n^2} + 1}}{{\sqrt {{n^4} + {n^2} + 1} + {n^2}}}\\
= \lim \frac{{1 + \frac{1}{{{n^2}}}}}{{\sqrt {1 + \frac{1}{{{n^2}}} + \frac{1}{{{n^4}}}} + 1}} = \frac{1}{2}
\end{array}\)
d. \(\lim \frac{1}{{\sqrt {{n^2} + 2n} - n}} = \lim \frac{{\sqrt {{n^2} + 2n} + n}}{{2n}}\)
\( = \lim \frac{{\sqrt {1 + \frac{2}{n}} + 1}}{2} = 1\)
-- Mod Toán 11
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