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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. \(\lim {1 \over {\sqrt {{n^2} + 2n} – n}}\)
a. \(\lim \sqrt {3{n^4} – 10n + 12} = \lim {n^2}.\sqrt {3 – {{10} \over {{n^3}}} + {{12} \over {{n^4}}}} \)
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\(= + \infty \)
b. \(\lim \left( {{{2.3}^n} – {{5.4}^n}} \right) = \lim {4^n}\left[ {2{{\left( {{3 \over 4}} \right)}^n} – 5} \right] = – \infty \)
c.
\(\eqalign{ & \lim \left( {\sqrt {{n^4} + {n^2} + 1} – {n^2}} \right) \cr&= \lim {{{n^2} + 1} \over {\sqrt {{n^4} + {n^2} + 1} + {n^2}}} \cr & = \lim {{1 + {1 \over {{n^2}}}} \over {\sqrt {1 + {1 \over {{n^2}}} + {1 \over {{n^4}}}} + 1}} = {1 \over 2} \cr} \)
d. \(\lim {1 \over {\sqrt {{n^2} + 2n }- n }} = \lim {{\sqrt {{n^2} + 2n} + n} \over {2n}} = \lim {{\sqrt {1 + {2 \over n} }+ 1 } \over 2} = 1\)