Tìm giới hạn:
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[4]{x}-1}{x^3+x-2}\)
K
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TV
Tử Văn Diệp
6 tháng 12 2023
Đúng(2)
Những câu hỏi liên quan
HT
24 tháng 1 2021
a/ L'Hospital:
\(=\lim\limits_{x\rightarrow2}\dfrac{x-\left(x+2\right)^{\dfrac{1}{2}}}{\left(4x+1\right)^{\dfrac{1}{2}}-3}=\lim\limits_{x\rightarrow2}\dfrac{1-\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}{\dfrac{1}{2}\left(4x+1\right)^{-\dfrac{1}{2}}.4}=\dfrac{1-\dfrac{1}{2}.4^{-\dfrac{1}{2}}}{2.9^{-\dfrac{1}{2}}}=\dfrac{9}{8}\)
b/ L'Hospital:\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7\right)^{\dfrac{1}{2}}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{2}\left(2x+7\right)^{-\dfrac{1}{2}}.2+1}{3x^2-8x}=\dfrac{9^{-\dfrac{1}{2}}+1}{3-8}=-\dfrac{4}{15}\)
NK
1
16 tháng 12 2023
\(\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x}+\sqrt{x-1}-1}{\sqrt{x^2-1}}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{\dfrac{\left(x-1\right)}{\sqrt{x}+1}+\left(\sqrt{x-1}\right)}{\sqrt{\left(x-1\right)\left(x+1\right)}}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{\left(\sqrt{x-1}\right)\left(\dfrac{\sqrt{x-1}}{\sqrt{x}+1}+1\right)}{\sqrt{x-1}\cdot\sqrt{x+1}}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{\left(\dfrac{\sqrt{x-1}}{\sqrt{x}+1}+1\right)}{\sqrt{x+1}}=\dfrac{\dfrac{\sqrt{1-1}}{\sqrt{1}+1}+1}{\sqrt{1+1}}\)
\(=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
30 tháng 12 2022
a) \(lim\dfrac{-2n+1}{n}=lim\dfrac{\dfrac{-2n}{n}+\dfrac{1}{n}}{\dfrac{n}{n}}=lim\dfrac{-2+\dfrac{1}{n}}{1}=\dfrac{lim\left(-2\right)+\dfrac{lim1}{n}}{lim1}=\dfrac{-2+0}{1}=-\dfrac{2}{1}=-2\)
b) \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{9-\left(x+8\right)}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{3+\sqrt{x+8}}=\dfrac{1}{3+\sqrt{1+8}}=\dfrac{1}{3+3}=\dfrac{1}{9}\)
AH
Akai Haruma
Giáo viên
3 tháng 4 2022
Lời giải:
\(L=\lim\limits_{x\to 1}\frac{\sqrt{2x-1}(\sqrt[3]{x+7}-2)+2(\sqrt{2x-1}-1)}{x(x-1)}=\lim\limits_{x\to 1}\frac{\sqrt{2x-1}.\frac{1}{\sqrt[3]{(x+7)^2}+2\sqrt[3]{x+7}+4}+4.\frac{1}{\sqrt{2x-1}+1}}{x}=\frac{25}{12}\)
QA
1
TH
1
HT
9 tháng 2 2021
\(=\lim\limits_{x\rightarrow1}\dfrac{2x-\dfrac{1}{2}.x^{-\dfrac{1}{2}}}{\dfrac{1}{2}.x^{-\dfrac{1}{2}}}=\dfrac{2-\dfrac{1}{2}}{\dfrac{1}{2}}=3\)
4 tháng 12 2023
a: \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-\sqrt{5-x^2}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-2+2-\sqrt{5-x^2}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x+7-8}{\sqrt[3]{\left(x+7\right)^2}+2\cdot\sqrt[3]{x+7}+4}+\dfrac{4-5+x^2}{2+\sqrt{5-x^2}}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\cdot\sqrt[3]{x+7}+4}+\dfrac{x^2-1}{2+\sqrt{5-x^2}}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(\dfrac{1}{\sqrt[3]{\left(x+7\right)^2}+2\cdot\sqrt[3]{\left(x+7\right)}+4}+\dfrac{x+1}{2+\sqrt{5-x^2}}\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{1}{\sqrt[3]{\left(x+7\right)^2}+2\cdot\sqrt[3]{x+7}+4}+\dfrac{x+1}{2+\sqrt{5-x^2}}\)
\(=\dfrac{1}{\sqrt[3]{\left(1+7\right)^2}+2\cdot\sqrt[3]{1+7}+4}+\dfrac{1+1}{2+\sqrt{5-1^2}}\)
\(=\dfrac{1}{4+2\cdot2+4}+\dfrac{2}{2+2}\)
\(=\dfrac{1}{12}+\dfrac{1}{2}=\dfrac{7}{12}\)
b: \(\lim\limits_{x\rightarrow5}\dfrac{x-5}{\sqrt{x}-\sqrt{5}}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\left(\sqrt{x}-\sqrt{5}\right)\left(\sqrt{x}+\sqrt{5}\right)}{\sqrt{x}-\sqrt{5}}\)
\(=\lim\limits_{x\rightarrow5}\sqrt{x}+\sqrt{5}=\sqrt{5}+\sqrt{5}=2\sqrt{5}\)