Tính các giới hạn
\(\lim\limits_{x\rightarrow1}\dfrac{x^2-\sqrt{x}}{\sqrt{x}-1}\)
K
Khách
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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:\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7\right)^{\dfrac{1}{2}}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{2}\left(2x+7\right)^{-\dfrac{1}{2}}.2+1}{3x^2-8x}=\dfrac{9^{-\dfrac{1}{2}}+1}{3-8}=-\dfrac{4}{15}\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 2 2021
Chúng ta tính giới hạn sau:
\(\lim\limits_{x\rightarrow1}\dfrac{1-\sqrt[n]{x}}{1-x}\)
Cách đơn giản nhất là sử dụng L'Hopital:
\(\lim\limits_{x\rightarrow1}\dfrac{1-x^{\dfrac{1}{n}}}{1-x}=\lim\limits_{x\rightarrow1}\dfrac{-\dfrac{1}{n}x^{\dfrac{1}{n}-1}}{-1}=\dfrac{1}{n}\)
Phức tạp hơn thì tách mẫu theo hằng đẳng thức
\(=\lim\limits_{x\rightarrow1}\dfrac{1-\sqrt[x]{n}}{\left(1-\sqrt[n]{x}\right)\left(1+\sqrt[n]{x}+\sqrt[n]{x^2}+...+\sqrt[n]{x^{n-1}}\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{1+\sqrt[n]{x}+\sqrt[n]{x^2}+...+\sqrt[n]{x^{n-1}}}=\dfrac{1}{n}\)
Tóm lại ta có:
\(\lim\limits_{x\rightarrow1}\dfrac{1-\sqrt[n]{x}}{1-x}=\dfrac{1}{n}\)
Do đó:
\(I_1=\lim\limits_{x\rightarrow1}\left(\dfrac{1-\sqrt[2]{x}}{1-x}\right)\left(\dfrac{1-\sqrt[3]{x}}{1-x}\right)...\left(\dfrac{1-\sqrt[n]{x}}{1-x}\right)=\dfrac{1}{2}.\dfrac{1}{3}...\dfrac{1}{n}=\dfrac{1}{n!}\)
Câu 2 cũng vậy: L'Hopital hoặc tách hằng đẳng thức trâu bò (thôi L'Hopital đi cho đỡ sợ)
\(I_2=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+x^2}+x\right)^n-\left(\sqrt{1+x^2}-x\right)^n}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{n\left(\sqrt{1+x^2}+x\right)^{n-1}\left(\dfrac{x}{\sqrt{1+x^2}}+1\right)-n\left(\sqrt{1+x^2}-x\right)^{n-1}\left(\dfrac{x}{\sqrt{1+x^2}}-1\right)}{1}\)
\(=\dfrac{n.1-n\left(-1\right)}{1}=2n\)
AH
Akai Haruma
Giáo viên
3 tháng 4 2022
Lời giải:
\(L=\lim\limits_{x\to 1}\frac{\sqrt{2x-1}(\sqrt[3]{x+7}-2)+2(\sqrt{2x-1}-1)}{x(x-1)}=\lim\limits_{x\to 1}\frac{\sqrt{2x-1}.\frac{1}{\sqrt[3]{(x+7)^2}+2\sqrt[3]{x+7}+4}+4.\frac{1}{\sqrt{2x-1}+1}}{x}=\frac{25}{12}\)
NK
1
16 tháng 12 2023
\(\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x}+\sqrt{x-1}-1}{\sqrt{x^2-1}}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{\dfrac{\left(x-1\right)}{\sqrt{x}+1}+\left(\sqrt{x-1}\right)}{\sqrt{\left(x-1\right)\left(x+1\right)}}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{\left(\sqrt{x-1}\right)\left(\dfrac{\sqrt{x-1}}{\sqrt{x}+1}+1\right)}{\sqrt{x-1}\cdot\sqrt{x+1}}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{\left(\dfrac{\sqrt{x-1}}{\sqrt{x}+1}+1\right)}{\sqrt{x+1}}=\dfrac{\dfrac{\sqrt{1-1}}{\sqrt{1}+1}+1}{\sqrt{1+1}}\)
\(=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
LH
1
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 1 2021
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x^2+3x+1\right)\left(\sqrt{3x+1}-2\right)+2\left(x^2+3x+1\right)-10}{x^2-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{3\left(x-1\right)\left(x^2+3x+1\right)}{\sqrt{3x+1}+2}+2\left(x-1\right)\left(x+4\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{3\left(x^2+3x+1\right)}{\sqrt{3x+1}+2}+2\left(x+4\right)}{x+1}=...\)
HA
1
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) \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{9-\left(x+8\right)}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{3+\sqrt{x+8}}=\dfrac{1}{3+\sqrt{1+8}}=\dfrac{1}{3+3}=\dfrac{1}{9}\)
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: \(\lim\limits_{x\rightarrow2}\dfrac{x^2-2x}{\sqrt{x+2}-2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)}{\dfrac{x+2-4}{\sqrt{x+2}+2}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)\cdot\left(\sqrt{x+2}+2\right)}{x-2}\)
\(=\lim\limits_{x\rightarrow2}x\left(\sqrt{x+2}+2\right)\)
\(=2\cdot\left(\sqrt{2+2}+2\right)\)
\(=2\cdot4=8\)
\(=\lim\limits_{x\rightarrow1}\dfrac{2x-\dfrac{1}{2}.x^{-\dfrac{1}{2}}}{\dfrac{1}{2}.x^{-\dfrac{1}{2}}}=\dfrac{2-\dfrac{1}{2}}{\dfrac{1}{2}}=3\)