\(lim\dfrac{2^n.3^n-3^n.4^n+5^{2n}}{3^n.4^n-4^n.5^n+6^{2n}}\) là: 0 1 2 \(\dfrac{1}{2}\) Hướng dẫn giải: \(lim\dfrac{2^n.3^n-3^n.4^n+5^{2n}}{3^n.4^n-4^n.5^n+6^{2n}}\)\(=lim\dfrac{\left(\dfrac{2}{6}\right)^n.\left(\dfrac{3}{6}\right)^n-\left(\dfrac{3}{6}\right)^n.\left(\dfrac{4}{6}\right)^n+\left(\dfrac{5}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n.\left(\dfrac{4}{6}\right)^n-\left(\dfrac{4}{6}\right)^n.\left(\dfrac{5}{6}\right)^n+1}=0\).
\(lim\left(n^4-2n.sin3n+1\right)\) là: \(1\) \(-\infty\) \(+\infty\) \(10^5\) Hướng dẫn giải: \(lim\left(n^4-2n.sin3n+1\right)\) \(=limn^4\left(1-\dfrac{2nsin3n}{n^4}+\dfrac{1}{n^4}\right)=+\infty\).
\(lim\left(\sqrt{2n^2+2n-1}-\sqrt{2n^2-n-5}\right)\) là: \(\dfrac{3\sqrt{2}}{4}\) \(\dfrac{3\sqrt{2}}{2}\) \(\dfrac{3\sqrt{2}}{5}\) \(\dfrac{3\sqrt{2}}{10}\) Hướng dẫn giải: \(lim\left(\sqrt{2n^2+2n-1}-\sqrt{2n^2-n-5}\right)\) \(=lim\dfrac{3n+4}{\sqrt{2n^2+2n-1}+\sqrt{2n^2-n-5}}=\dfrac{3\sqrt{3}}{4}\).
Tìm \(lim\dfrac{\sqrt[4]{n^8-4n^4+2n}}{n^2+3n+1}\) 1 2 4 \(\dfrac{4}{3}\) Hướng dẫn giải: \(lim\dfrac{\sqrt[4]{n^8-4n^4+2n}}{n^2+3n+1}\) \(=lim\dfrac{n^2\sqrt[4]{1-\dfrac{4}{n^4}+\dfrac{2}{n^7}}}{n^2\left(1+\dfrac{3}{n}+\dfrac{1}{n^2}\right)}=1\).
Giới hạn nào trong bốn giới hạn sau đây có giá trị bằng 2 ? \(lim\left(2n-cosn\right)\) \(lim\dfrac{4n^2-2n.cos3n+2}{\sqrt{4n^4+2n^2+8}}\) \(lim\left(n^2+n-4\sqrt[3]{n}\right)\) \(lim\dfrac{3n^2+n}{n+2}\)
Tính \(\lim\left(\sqrt{2n^2-n+1}-\sqrt{2n^2-4n+5}\right).n\) \(+\infty\) \(-\infty\) \(1\) \(2\) Hướng dẫn giải: \(\left(\sqrt{2n^2-n+1}-\sqrt{2n^2-4n+5}\right).n\) \(=\dfrac{n\left(3n-4\right)}{\sqrt{2n^2-n+1}+\sqrt{2n^2-4n+5}}=n.\dfrac{3-\dfrac{4}{n}}{\sqrt{2-\dfrac{1}{n}+\dfrac{1}{n^2}}+\sqrt{2-\dfrac{4}{n}+\dfrac{5}{n^2}}}\)Giới hạn cần tính là \(+\infty\)
\(lim\dfrac{1}{\sqrt{4n+2}-\sqrt{2n+1}}\)là: 0 1 2 \(\dfrac{1}{2}\) Hướng dẫn giải: \(lim\dfrac{1}{\sqrt{4n+2}-\sqrt{2n+1}}\) \(=lim\dfrac{\sqrt{4n+2}+\sqrt{2n-1}}{2n-1}\) \(=lim\dfrac{\sqrt{\dfrac{4n}{n^2}+\dfrac{1}{n^2}}+\sqrt{\dfrac{2n}{n^2}-\dfrac{1}{n^2}}}{2-\dfrac{1}{n^2}}\) \(=0\).
Tính \(\lim\dfrac{\left(2n+1\right)\left(n^2+3n\right)}{\left(n+1\right)^2.n}\) là: 2 3 1 4 Hướng dẫn giải: Có \(\dfrac{\left(2n+1\right)\left(n^2+3n\right)}{\left(n+1\right)^2.n}\text{}\)\(=\dfrac{\left(2-\dfrac{1}{n}\right)\left(1+\dfrac{3}{n}\right)}{\left(1+\dfrac{1}{n}\right)^2.1}\) nên \(\lim\dfrac{\left(2n+1\right)\left(n^2+3n\right)}{\left(n+1\right)^2.n}=\dfrac{\left(2-0\right)\left(1+0\right)}{\left(1+0\right)^2}=2.\)
Tính \(\lim(\sqrt{3n.3^n-3^n}+8)\) . 1 2 3 \(+\infty\) Hướng dẫn giải: \(\lim(\sqrt{3n.3^n-3^n}+8)\) \(=\lim(3^{\dfrac{n}{2}}\sqrt{\left(3n-1\right)}+8)=+\infty\).
