Tìm giới hạn : E=lim $\frac{ \sqrt{n^{3}+2n }+1}{n+2}$ B= lim( $\sqrt[3]{n^{3}+9{n^{2}} }$-n) N=lim( $\sqrt[3]{n^{3}+3n^{2}+1 }$ -n) C=lim( $\frac{3.

$\begin{array}{l}+)\quad E = \lim\dfrac{\sqrt{n^3 +2n} + 1}{n+2}\\
\to E = \lim\dfrac{n\sqrt{n + \dfrac{2}{n^2}} + 1}{n+2} \\
\to E = \lim\dfrac{\sqrt{n + \dfrac{2}{n^2}} + \dfrac1n}{1 + \dfrac2n}\\
\to E = \dfrac{\sqrt{+\infty + 0} + 0}{1+0}\\
\to E = +\infty\\
+)\quad B = \lim(\sqrt[3]{n^3 + 9n^2 } -n)\\
\to B= \lim\dfrac{(\sqrt[3]{n^3 + 9n^2 } -n)(\sqrt[3]{(n^3 + 9n^2)^2} + n\sqrt[3]{n^3 + 9n^2} + n^2)}{\sqrt[3]{(n^3 + 9n^2)^2} + n\sqrt[3]{n^3 + 9n^2} + n^2}\\
\to B = \lim\dfrac{9n^2}{\sqrt[3]{(n^3 + 9n^2)^2} + n\sqrt[3]{n^3 + 9n^2} + n^2}\\
\to B = \lim\dfrac{9}{\sqrt[3]{\left(1 + \dfrac9n\right)^2} + \sqrt[3]{1 + \dfrac9n} + 1}\\
\to B= \dfrac{9}{\sqrt[3]{(1 + 0)^2} + \sqrt[3]{1 +0} +1}\\
\to B = \dfrac{9}{1+1+1}\\
\to B = 3\\
+)\quad  N = \lim(\sqrt[3]{n^3+3n^2 – 1} -n)\\
\to N = \lim\dfrac{(\sqrt[3]{n^3+3n^2 – 1} -n)(\sqrt[3]{(n^3 + 3n^2 – 1)^2} + n\sqrt[3]{n^3 + 3n^2 -1} + n^2)}{\sqrt[3]{(n^3 + 3n^2 – 1)^2} + n\sqrt[3]{n^3 + 3n^2 -1} + n^2}\\
\to N = \lim\dfrac{3n^2 -1}{\sqrt[3]{(n^3 + 3n^2 – 1)^2} + n\sqrt[3]{n^3 + 3n^2 -1} + n^2}\\
\to N = \lim\dfrac{3 – \dfrac{1}{n^2}}{\sqrt[3]{\left(1 + \dfrac3n – \dfrac{1}{n^3}\right)^2} + \sqrt[3]{1 + \dfrac3n – \dfrac{1}{n^3}} + 1}\\
\to N = \dfrac{3-0}{\sqrt[3]{(1 + 0 -0)^2} + \sqrt[3]{1 + 0 -0} +1}\\
\to N = \dfrac{3}{1+1+1}\\
\to N = 1\\
+)\quad C =\lim\dfrac{3.2^n -3^n}{2^{n+1}+3^{n+1}}\\
\to C = \lim\dfrac{3.2^n -3^n}{2.2^{n}+3.3^{n}}\\
\to C = \lim\dfrac{3\cdot \left(\dfrac23\right)^n – 1}{2\cdot\left(\dfrac23\right)^n + 3}\\
\to C = \dfrac{3.0 – 1}{2.0 + 3}\\
\to C = -\dfrac13

\end{array}$