Câu 12:
\(lim\dfrac{\sqrt[3]{n^3+2023}}{2n+2024}\)
\(=\lim\limits\dfrac{\sqrt[3]{1+\dfrac{2023}{n^3}}}{2+\dfrac{2024}{n}}=\dfrac{\sqrt[3]{1}}{2}=\dfrac{1}{2}\)
=>Chọn B
Câu 13:
\(\lim\limits\dfrac{\sqrt{n+5}-2n}{n-1}=\lim\limits\dfrac{\sqrt{1+\dfrac{5}{n}}-2}{1-\dfrac{1}{n}}=\dfrac{0-2}{1-0}=-2\)
=>Chọn D
Câu 14:
\(\lim\limits\dfrac{3\sqrt{n+5}-2\sqrt{n}}{n^2+5}=\lim\limits\dfrac{3\sqrt{\dfrac{1}{n^3}+\dfrac{5}{n^4}}-2\sqrt{\dfrac{1}{n^3}}}{1+\dfrac{5}{n^2}}=0\)
=>Chọn A
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