Quay lại đáp án

1) $lim_{n \to +\infty} \left ( \sqrt[3]{n^3+n^2}-\sqrt{n^2-1}\right )=lim_{x \to +\infty}\left ( \sqrt[3]{n^3+n^2}-n + n-\sqrt{n^2-1} \right )$$=lim_{n \to + \infty}\left ( \frac{n^2}{\sqrt[3]{n^6 +2n^5 + n^4}+\sqrt[3]{n^6+n^5}+n^2} +\frac{1}{n+\sqrt{n^2-1}}\right )$$=lim_{n \to +\infty}\left ( \frac{1}{\sqrt[3]{1+\frac{2}{n}+\frac{1}{n^2}}+\sqrt[3]{1+\frac{1}{n}}+1}+\frac{\frac{1}{n}}{1+\sqrt{1-\frac{1}{n^2}}} \right )$$\Rightarrow lim_{n \to +\infty}\left ( .... \right )=\frac{1}{3}$

1) $lim_{n \to +\infty} \left ( \sqrt[3]{n^3+n^2}-\sqrt{n^2-1}\right )=lim_{n \to +\infty}\left ( \sqrt[3]{n^3+n^2}-n + n-\sqrt{n^2-1} \right )$$=lim_{n \to + \infty}\left ( \frac{n^2}{\sqrt[3]{n^6 +2n^5 + n^4}+\sqrt[3]{n^6+n^5}+n^2} +\frac{1}{n+\sqrt{n^2-1}}\right )$$=lim_{n \to +\infty}\left ( \frac{1}{\sqrt[3]{1+\frac{2}{n}+\frac{1}{n^2}}+\sqrt[3]{1+\frac{1}{n}}+1}+\frac{\frac{1}{n}}{1+\sqrt{1-\frac{1}{n^2}}} \right )$$\Rightarrow lim_{n \to +\infty}\left ( .... \right )=\frac{1}{3}$

1) $lim_{x \to +\infty} \left ( \sqrt[3]{n^3+n^2}-\sqrt{n^2-1}\right )=lim_{x \to +\infty}\left ( \sqrt[3]{n^3+n^2}-n + n-\sqrt{n^2-1} \right )$$=lim_{x \to + \infty}\left ( \frac{n^2}{\sqrt[3]{n^6 +2n^5 + n^4}+\sqrt[3]{n^6+n^5}+n^2} +\frac{1}{n+\sqrt{n^2-1}}\right )$$=lim_{x \to +\infty}\left ( \frac{1}{\sqrt[3]{1+\frac{2}{n}+\frac{1}{n^2}}+\sqrt[3]{1+\frac{1}{n}}+1}+\frac{\frac{1}{n}}{1+\sqrt{1-\frac{1}{n^2}}} \right )$$\Rightarrow lim_{x \to +\infty}\left ( .... \right )=\frac{1}{3}$

1) $lim_{n \to +\infty} \left ( \sqrt[3]{n^3+n^2}-\sqrt{n^2-1}\right )=lim_{x \to +\infty}\left ( \sqrt[3]{n^3+n^2}-n + n-\sqrt{n^2-1} \right )$$=lim_{n \to + \infty}\left ( \frac{n^2}{\sqrt[3]{n^6 +2n^5 + n^4}+\sqrt[3]{n^6+n^5}+n^2} +\frac{1}{n+\sqrt{n^2-1}}\right )$$=lim_{n \to +\infty}\left ( \frac{1}{\sqrt[3]{1+\frac{2}{n}+\frac{1}{n^2}}+\sqrt[3]{1+\frac{1}{n}}+1}+\frac{\frac{1}{n}}{1+\sqrt{1-\frac{1}{n^2}}} \right )$$\Rightarrow lim_{n \to +\infty}\left ( .... \right )=\frac{1}{3}$

1) $lim_{x \to +\infty} \left ( \sqrt[3]{n^3+n^2}-\sqrt{n^2-1}\right )=lim_{x \to +\infty}\left ( \sqrt[3]{n^3+n^2}-n + n-\sqrt{n^2-1} \right )$$=lim_{x \to + \infty}\left ( \frac{n^2}{\sqrt[3]{n^6 +2n^5 + n^4}+\sqrt[3]{n^6+n^5}+n^2} +\frac{1}{n+\sqrt{n^2-1}}\right )$$=lim_{x \to +\infty}\left ( \frac{1}{\sqrt[3]{1+\frac{2}{n}+\frac{1}{n^2}}}+\frac{\frac{1}{n}}{1+\sqrt{1-\frac{1}{n^2}}} \right )$$\Rightarrow lim_{x \to +\infty}\left ( .... \right )=1$

1) $lim_{x \to +\infty} \left ( \sqrt[3]{n^3+n^2}-\sqrt{n^2-1}\right )=lim_{x \to +\infty}\left ( \sqrt[3]{n^3+n^2}-n + n-\sqrt{n^2-1} \right )$$=lim_{x \to + \infty}\left ( \frac{n^2}{\sqrt[3]{n^6 +2n^5 + n^4}+\sqrt[3]{n^6+n^5}+n^2} +\frac{1}{n+\sqrt{n^2-1}}\right )$$=lim_{x \to +\infty}\left ( \frac{1}{\sqrt[3]{1+\frac{2}{n}+\frac{1}{n^2}}+\sqrt[3]{1+\frac{1}{n}}+1}+\frac{\frac{1}{n}}{1+\sqrt{1-\frac{1}{n^2}}} \right )$$\Rightarrow lim_{x \to +\infty}\left ( .... \right )=\frac{1}{3}$