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( $\frac{3.2^{n}-3^{n}}{2^{n+1}+3^{n+1}}$ )
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.
By Audrey
Đáp án:
Giải thích các bước giải:
$\begin{array}{l}+)\quad E = \lim\dfrac{\sqrt{n^3 +2n} + 1}{n+2}\\ \to E = \lim\dfrac{n\sqrt{n + \dfrac{2}{n^2}} + 1}{n+2} \\ \to E = \lim\dfrac{\sqrt{n + \dfrac{2}{n^2}} + \dfrac1n}{1 + \dfrac2n}\\ \to E = \dfrac{\sqrt{+\infty + 0} + 0}{1+0}\\ \to E = +\infty\\ +)\quad B = \lim(\sqrt[3]{n^3 + 9n^2 } -n)\\ \to B= \lim\dfrac{(\sqrt[3]{n^3 + 9n^2 } -n)(\sqrt[3]{(n^3 + 9n^2)^2} + n\sqrt[3]{n^3 + 9n^2} + n^2)}{\sqrt[3]{(n^3 + 9n^2)^2} + n\sqrt[3]{n^3 + 9n^2} + n^2}\\ \to B = \lim\dfrac{9n^2}{\sqrt[3]{(n^3 + 9n^2)^2} + n\sqrt[3]{n^3 + 9n^2} + n^2}\\ \to B = \lim\dfrac{9}{\sqrt[3]{\left(1 + \dfrac9n\right)^2} + \sqrt[3]{1 + \dfrac9n} + 1}\\ \to B= \dfrac{9}{\sqrt[3]{(1 + 0)^2} + \sqrt[3]{1 +0} +1}\\ \to B = \dfrac{9}{1+1+1}\\ \to B = 3\\ +)\quad N = \lim(\sqrt[3]{n^3+3n^2 – 1} -n)\\ \to N = \lim\dfrac{(\sqrt[3]{n^3+3n^2 – 1} -n)(\sqrt[3]{(n^3 + 3n^2 – 1)^2} + n\sqrt[3]{n^3 + 3n^2 -1} + n^2)}{\sqrt[3]{(n^3 + 3n^2 – 1)^2} + n\sqrt[3]{n^3 + 3n^2 -1} + n^2}\\ \to N = \lim\dfrac{3n^2 -1}{\sqrt[3]{(n^3 + 3n^2 – 1)^2} + n\sqrt[3]{n^3 + 3n^2 -1} + n^2}\\ \to N = \lim\dfrac{3 – \dfrac{1}{n^2}}{\sqrt[3]{\left(1 + \dfrac3n – \dfrac{1}{n^3}\right)^2} + \sqrt[3]{1 + \dfrac3n – \dfrac{1}{n^3}} + 1}\\ \to N = \dfrac{3-0}{\sqrt[3]{(1 + 0 -0)^2} + \sqrt[3]{1 + 0 -0} +1}\\ \to N = \dfrac{3}{1+1+1}\\ \to N = 1\\ +)\quad C =\lim\dfrac{3.2^n -3^n}{2^{n+1}+3^{n+1}}\\ \to C = \lim\dfrac{3.2^n -3^n}{2.2^{n}+3.3^{n}}\\ \to C = \lim\dfrac{3\cdot \left(\dfrac23\right)^n – 1}{2\cdot\left(\dfrac23\right)^n + 3}\\ \to C = \dfrac{3.0 – 1}{2.0 + 3}\\ \to C = -\dfrac13 \end{array}$