tính giá trị của giới hạn \(\lim\limits_{x\rightarrow-\infty}\left(x+\sqrt{x^2+5x}\right)\)
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9 tháng 2 2021
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[n]{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)}-x\right)\\ =\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)-x^n}{\sqrt[n]{\left(\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)\right)^{n-1}}+...+x^{n-1}}\right)\)
= hệ số xn-1 trên tử/hệ số xn-1 dưới mẫu = \(\dfrac{a_1+a_2+...+a_n}{n}\)
19 tháng 2 2023
a: \(\lim\limits_{x->0^-^-}\dfrac{-2x+x}{x\left(x-1\right)}=lim_{x->0^-}\left(\dfrac{-x}{x\left(x-1\right)}\right)\)
\(=lim_{x->0^-}\left(\dfrac{-1}{x-1}\right)=\dfrac{-1}{0-1}=\dfrac{-1}{-1}=1\)
b: \(=lim_{x->-\infty}\left(\dfrac{x^2-x-x^2+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)
\(=lim_{x->-\infty}\left(\dfrac{-x+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)
\(=lim_{x->-\infty}\left(\dfrac{-1+\dfrac{1}{x}}{-\sqrt{1-\dfrac{1}{x^2}}-\sqrt{1-\dfrac{1}{x^2}}}\right)=\dfrac{-1}{-2}=\dfrac{1}{2}\)
HT
4 tháng 4 2021
Xet \(m\ne-3\)
\(=\lim\limits_{x\rightarrow-\infty}x\left(\sqrt[3]{1}+\sqrt{4}+m\right)=x\left(3+m\right)\)
\(=\left[{}\begin{matrix}-\infty\left(m>-3\right)\\+\infty\left(m< -3\right)\end{matrix}\right.\)
Xet \(m=-3\)
\(=\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3+2x^2+1}-x-2x-\sqrt{4x^2+2x+3}\right)\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^3+2x^2+1-x^3}{\sqrt[3]{\left(x^3+2x^2+1\right)^2}+x\sqrt[3]{x^3+2x^2+1}+x^2}-\lim\limits_{x\rightarrow-\infty}\dfrac{4x^2-4x^2-2x-3}{2x-\sqrt{4x^2+2x+3}}\)
\(=\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{7}{6}\)
AH
Akai Haruma
Giáo viên
4 tháng 4 2021
Bạn bị nhầm số rồi. Xét $m>1; m< 1; m=1$ mới đúng chứ
Y
15 tháng 10 2023
\(1,\lim\limits_{n\rightarrow\infty}\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}\left(1\right)\)
\(\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}=\dfrac{-\dfrac{n^2}{n^4}+\dfrac{2n}{n^4}+\dfrac{1}{n^4}}{\sqrt{\dfrac{3n^4}{n^4}+\dfrac{2}{n^4}}}=\dfrac{-\dfrac{1}{n^2}+\dfrac{2}{n^3}+\dfrac{1}{n^4}}{\sqrt{3+\dfrac{2}{n^4}}}\)
\(\Rightarrow\left(1\right)=\dfrac{-lim\dfrac{1}{n^2}+2lim\dfrac{1}{n^3}+lim\dfrac{1}{n^4}}{\sqrt{lim\left(3+\dfrac{2}{n^4}\right)}}\)
\(=\dfrac{0}{\sqrt{lim\left(3+\dfrac{2}{n^4}\right)}}=0\)
\(2,\lim\limits_{n\rightarrow\infty}\left(\dfrac{4n-\sqrt{16n^2+1}}{n+1}\right)\left(2\right)\)
\(\dfrac{4n-\sqrt{16n^2+1}}{n+1}=\dfrac{\dfrac{4n}{n^2}-\sqrt{\dfrac{16n^2}{n^2}+\dfrac{1}{n^2}}}{\dfrac{n}{n^2}+\dfrac{1}{n^2}}=\dfrac{\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}}{\dfrac{1}{n}+\dfrac{1}{n^2}}\)
\(\Rightarrow\left(2\right)=\dfrac{lim\left(\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}\right)}{lim\left(\dfrac{1}{n}+\dfrac{1}{n^2}\right)}=\dfrac{lim\left(\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}\right)}{0}\)
Vậy giới hạn \(\left(2\right)\) không xác định.
\(3,\lim\limits_{n\rightarrow\infty}\left(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}\right)\left(3\right)\)
\(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}=\dfrac{\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}}{\dfrac{2}{n}}\)
\(\Rightarrow\left(3\right)=\dfrac{lim\left(\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}\right)}{2lim\dfrac{1}{n}}=\dfrac{lim\left(\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}\right)}{0}\)
Vậy \(lim\left(3\right)\) không xác định.
HT
27 tháng 1 2021
a/ \(=\lim\limits_{x\rightarrow-\infty}x^3\left(3+\dfrac{5x^2}{x^3}-\dfrac{9\sqrt{2}x}{x^3}-\dfrac{2017}{x^3}\right)=3.x^3=-\infty\)
b/ \(=\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{1+\dfrac{x}{x^2}+\dfrac{1}{x^2}}-\sqrt[3]{2+\dfrac{x}{x^3}-\dfrac{1}{x^3}}\right)=\left(1-\sqrt[3]{2}\right)x=-\infty\)
c/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2-x^2-x-1}{x+\sqrt{x^2+x+1}}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\dfrac{x}{x}-\dfrac{1}{x}}{\dfrac{x}{x}-\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}}=-\dfrac{1}{1-1}=-\infty\)
d/ \(=\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3+x^2+1}-x\right)+\lim\limits_{x\rightarrow-\infty}\left(x+\sqrt{x^2+x+1}\right)\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^3+x^2+1-x^3}{\left(\sqrt[3]{x^3+x^2+1}\right)^2+x\sqrt[3]{x^3+x^2+1}-x^2}+\lim\limits_{x\rightarrow-\infty}\dfrac{x^2-x^2-x-1}{x-\sqrt{x^2+x+1}}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+1}{\left(-x\sqrt[3]{\dfrac{x^3}{x^3}+\dfrac{x^2}{x^3}+\dfrac{1}{x^3}}\right)^2-x.x\sqrt[3]{\dfrac{x^3}{x^3}+\dfrac{x^2}{x^3}+\dfrac{1}{x^3}}-x^2}+\lim\limits_{x\rightarrow-\infty}\dfrac{-x-1}{x+x\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}}\)
\(=\dfrac{1}{1-1-1}+\dfrac{-1}{1+1}=-1-\dfrac{1}{2}=-\dfrac{3}{2}\)
\(\lim\limits_{x\rightarrow-\infty}\left(x+\sqrt{x^2+5x}\right)=\lim\limits_{x\rightarrow-\infty}\dfrac{-5x}{x-\sqrt{x^2+5x}}\\ =\lim\limits_{x\rightarrow-\infty}\dfrac{5}{-1-\sqrt{1+\dfrac{5}{x}}}=-\dfrac{5}{2}\)
Quá sai.