\(\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x^2-3x+2}}{x^2-5x+4}\)
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NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 2 2021
\(a=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-1+1-\sqrt[3]{2x+1}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x}{\sqrt[]{4x+1}+1}+\dfrac{-2x}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{4x+1}+1}+\dfrac{-2}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}\right)=...\)
\(b=\lim\limits_{x\rightarrow1}\dfrac{4\left(x-1\right)\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(x-1\right)\left(\sqrt[]{4x+5}+3\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{4\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(\sqrt[]{4x+5}+3\right)}=...\)
\(c=\lim\limits_{x\rightarrow-1}\dfrac{\left(2x+3\right)^{\dfrac{1}{4}}+\left(2+3x\right)^{\dfrac{1}{3}}}{\left(x+2\right)^{\dfrac{1}{2}}-1}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{1}{2}\left(2x+3\right)^{-\dfrac{3}{4}}+\left(2+3x\right)^{-\dfrac{2}{3}}}{\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}=3\)
C
1
RH
26 tháng 12 2023
\(\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{x^2-3x+2}}{x^2-5x+4}=\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{\left(1-x\right)\left(2-x\right)}}{\left(1-x\right)\left(4-x\right)}\\ =\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{2-x}}{\left(4-x\right)\sqrt{1-x}}\)
\(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow1^-}\sqrt{2-x}=1>0\\\lim\limits_{x\rightarrow1^-}\left(4-x\right)\sqrt{1-x}=0\\x< 1\rightarrow\left(4-x\right)\sqrt{1-x}>0\end{matrix}\right.\\ \rightarrow\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{x^2-3x+2}}{x^2-5x+4}=\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{2-x}}{\left(4-x\right)\sqrt{1-x}}=+\infty\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 1 2021
\(a=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x^2-2x-2\right)}{\left(x-1\right)\left(x-3\right)}=\lim\limits_{x\rightarrow1}\dfrac{x^2-2x-2}{x-3}=\dfrac{3}{2}\)
Câu b bạn coi lại đề, là \(x\rightarrow-1^-\) hay \(x\rightarrow1^-\) (đúng như đề thì ko phải dạng vô định, cứ thay số rồi bấm máy)
\(c=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)}{\left(x-3\right)\left(x-1\right)\left(\sqrt[3]{\left(x+5\right)^2}+2\sqrt[3]{x+5}+4\right)}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{1}{\left(x-1\right)\left(\sqrt[3]{\left(x+5\right)^2}+2\sqrt[3]{x+5}+4\right)}=\dfrac{1}{2.\left(4+4+4\right)}=...\)
HT
27 tháng 1 2021
a/ \(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}{\left(x-1\right)\left(x-3\right)}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}{x-3}=....\)
Từ 2 câu kia lát tui làm, ăn cơm đã :D
HT
7 tháng 2 2021
1/ \(=\lim\limits_{x\rightarrow0}\dfrac{3\left(1+3x\right)^2.3+4.4\left(1-4x\right)^3}{1}=...\left(thay-x-vo\right)\)
2/ \(=\lim\limits_{x\rightarrow2}\dfrac{2.2.x-5}{3x^2-3}=\dfrac{4.2-5}{3.4-3}=\dfrac{1}{3}\)
3/ \(=\lim\limits_{x\rightarrow1}\dfrac{4x^3-3}{3x^2+2}=\dfrac{4.1-3}{3.1-2}=1\)
Xai L'Hospital nhe :v
23 tháng 2 2021
câu 1 bạn lm kiểu j vậy chả hiểu luôn bạn có thể lm lại chi tiết hơn dc ko
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 3 2022
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+2}+\sqrt{5x+4}-5}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+2}-2+\sqrt{5x+4}-3}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{2\left(x-1\right)}{\sqrt{2x+2}+2}+\dfrac{5\left(x-1\right)}{\sqrt{5x+4}+3}}{x-1}=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\sqrt{2x+2}+2}+\dfrac{5}{\sqrt{5x+4}+3}\right)=\dfrac{2}{2+2}+\dfrac{5}{3+3}=...\)
Đề câu b là \(...\sqrt{90-6x}\) hay \(\sqrt{9-6x}\) vậy em? Hình như cái sau mới có lý
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 2 2019
1/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}.\sqrt[4]{1+8x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}-1}{x}\)
Liên hợp dài quá ko muốn gõ tiếp, bạn tự đặt nhân tử chung rồi liên hợp nhé, kết quả ra 5
2/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{1+7x}-2-\left(x^3-3x+2\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{7\left(x-1\right)}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)^2\left(x+2\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{7}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)\left(x+2\right)=\dfrac{7}{12}\)
3/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{x^3-x^2+1}{2x^2+3x-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{x-1+\dfrac{1}{x^2}}{2+\dfrac{3}{x}-\dfrac{1}{x^2}}=-\infty\)
4/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}}{\sqrt{4x+1}}=\lim\limits_{x\rightarrow+\infty}\dfrac{1+\dfrac{1}{\sqrt[6]{x}}+\dfrac{1}{\sqrt[4]{x}}}{\sqrt{4+\dfrac{1}{x}}}=\dfrac{1}{\sqrt{4}}=\dfrac{1}{2}\)
5/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+\sqrt{x^2+2}}{\sqrt[3]{8x^3+x^2+1}}=\lim\limits_{x\rightarrow-\infty}\dfrac{1-\sqrt{1+\dfrac{2}{x^2}}}{\sqrt[3]{8+\dfrac{1}{x}+\dfrac{1}{x^3}}}=\dfrac{1-1}{\sqrt[3]{8}}=0\)
6/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+3x-7}}{\sqrt[3]{27x^3+5x^2+x-4}}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{4+\dfrac{3}{x}-\dfrac{7}{x^2}}}{\sqrt[3]{27+\dfrac{5}{x}+\dfrac{1}{x^2}-\dfrac{4}{x^3}}}=\dfrac{-\sqrt{4}}{\sqrt[3]{27}}=\dfrac{-2}{3}\)
HT
27 tháng 1 2021
a/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}\sqrt{x^2+1}+\dfrac{2x}{x}+\dfrac{1}{x}}{\dfrac{x}{x}\sqrt[3]{\dfrac{2x^3}{x^3}+\dfrac{x}{x^3}+\dfrac{1}{x^3}}+\dfrac{x}{x}}=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+1}+2}{\sqrt[3]{2}+1}=+\infty\)
b/ \(=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2.1^2-1+1}-\sqrt[3]{2.1+3}}{3.1^2-2}=...\)
c/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+x\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{x\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)
10 tháng 11 2023
a: \(\lim\limits_{x\rightarrow1}\dfrac{x^2-1}{\sqrt{3x+1}-2}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)}{\dfrac{3x+1-4}{\sqrt{3x+1}+2}}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)\cdot\left(\sqrt{3x+1}+2\right)}{3\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x+1\right)\left(\sqrt{3x+1}+2\right)}{3}\)
\(=\dfrac{\left(1+1\right)\left(\sqrt{3+1}+2\right)}{2}=\dfrac{2\cdot4}{3}=\dfrac{8}{3}\)
b: \(\lim\limits_{x\rightarrow2}\dfrac{x^2-2x}{\sqrt{x+2}-2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)}{\dfrac{x+2-4}{\sqrt{x+2}+2}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)\cdot\left(\sqrt{x+2}+2\right)}{x-2}\)
\(=\lim\limits_{x\rightarrow2}x\left(\sqrt{x+2}+2\right)\)
\(=2\cdot\left(\sqrt{2+2}+2\right)\)
\(=2\cdot4=8\)
Không tồn tại.