Tính \(lim_{x\rightarrow1}\dfrac{\sqrt{2x+7}-3}{x^3-2x^2+2022x-2021}\)
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HT
Hoàng Tử Hà
18 tháng 5 2021
Đúng(1)
Những câu hỏi liên quan
AH
Akai Haruma
Giáo viên
3 tháng 4 2022
Lời giải:
\(L=\lim\limits_{x\to 1}\frac{\sqrt{2x-1}(\sqrt[3]{x+7}-2)+2(\sqrt{2x-1}-1)}{x(x-1)}=\lim\limits_{x\to 1}\frac{\sqrt{2x-1}.\frac{1}{\sqrt[3]{(x+7)^2}+2\sqrt[3]{x+7}+4}+4.\frac{1}{\sqrt{2x-1}+1}}{x}=\frac{25}{12}\)
30 tháng 10 2021
\(a,\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\\ \Rightarrow2x=-4\Rightarrow x=-2\\ b,\Rightarrow\left(x-2021\right)\left(2022x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{2022}\end{matrix}\right.\\ c,\Rightarrow\left(x^2-9\right)-\left(x-3\right)\left(2x+7\right)=0\\ \Rightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(2x+7\right)=0\\ \Rightarrow\left(x-3\right)\left(x+3-2x-7\right)=0\\ \Rightarrow\left(x-3\right)\left(-4-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
27 tháng 11 2023
Lời giải:
\(\lim\limits_{x\to 1-}\frac{2x+1}{x-1}=-\infty\) do với $x\to 1-$ thì $\lim(2x+1)=3>0$ và $\lim (x-1)=0$ và $x-1<0$
\(\lim\limits_{x\to 6}\frac{(5x-4)\sqrt{2x-3}+x-84}{x-6}=\lim\limits_{x\to 6}\frac{(5x-4)(\sqrt{2x-3}-3)+16(x-6)}{x-6}\)
\(=\lim\limits_{x\to 6}\frac{(5x-4).\frac{2(x-6)}{\sqrt{2x-3}+3}+16(x-6)}{x-6}=\lim\limits_{x\to 6}[\frac{2(5x-4)}{\sqrt{2x-3}+3}+16]=\frac{74}{3}\)
RH
26 tháng 12 2023
\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-2x+1}{x-1}=3\rightarrow\lim\limits_{x\rightarrow1}\left(f\left(x\right)-2x+1\right)=0\\ \rightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=1\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{3f\left(x\right)+1}-x-1}{\sqrt{4x+5}-3x-2}=\dfrac{\sqrt{3.1+1}-1-1}{\sqrt{4.1+5}-3.1-2}=0\)
TK
1
24 tháng 11 2023
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{4x+5}-2x-3}{\left(x+1\right)^2}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{4x+5-\left(2x+3\right)^2}{\sqrt{4x+5}+2x+3}\cdot\dfrac{1}{\left(x+1\right)^2}\)
\(=\lim\limits_{x\rightarrow-1}\left(\dfrac{4x+5-4x^2-12x-9}{\left(\sqrt{4x+5}+2x+3\right)\cdot\left(x+1\right)^2}\right)\)
\(=\lim\limits_{x\rightarrow-1}\left(\dfrac{-4x^2-8x-4}{\left(\sqrt{4x+5}+2x+3\right)\cdot\left(x+1\right)^2}\right)\)
\(=\lim\limits_{x\rightarrow-1}\left(\dfrac{-4\left(x^2+2x+1\right)}{\left(x+1\right)^2\cdot\left(\sqrt{4x+5}+2x+3\right)}\right)\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{-4}{\sqrt{4x+5}+2x+3}\)
\(=\dfrac{-4}{\sqrt{-4+5}-2+3}=\dfrac{-4}{1+1}=-\dfrac{4}{2}=-2\)
HT
24 tháng 1 2021
a/ \(=\lim\limits_{h\rightarrow0}\dfrac{2x^3+6x^2h+6xh^2+2h^3-2x^3}{h}\)
\(=\lim\limits_{h\rightarrow0}\dfrac{6xh^2+6x^2h+2h^3}{h}=\lim\limits_{h\rightarrow0}\left(6xh+6x^2+2h^2\right)=6x^2\)
b/ Xet day :\(S=x+x^2+....+x^{2021}\)
Day co \(\left\{{}\begin{matrix}u_1=x\\q=x\end{matrix}\right.\Rightarrow S=u_1.\dfrac{q^{2021}-1}{q-1}=x.\dfrac{x^{2021}-1}{x-1}\)
\(\Rightarrow\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^{2022}-x}{x-1}-2021}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{x^{2022}-x-2021x+2021}{\left(x-1\right)^2}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^{2022}}{x^2}-\dfrac{x}{x^2}-\dfrac{2021x}{x^2}+\dfrac{2021}{x^2}}{\dfrac{x^2}{x^2}-\dfrac{2x}{x^2}+\dfrac{1}{x^2}}=\lim\limits_{x\rightarrow1}\dfrac{x^{2020}}{1}=1\)
HT
24 tháng 1 2021
Lam lai cau b, hinh nhu bi nham sang dang \(\dfrac{\infty}{\infty}\) roi
Xet day: \(S=x+x^2+...+x^{2021}\)
\(\Rightarrow S=x.\dfrac{x^{2021}-1}{x-1}=\dfrac{x^{2022}-x}{x-1}\)
\(\Rightarrow\lim\limits_{x\rightarrow1}\dfrac{x^{2022}-2022x+2021}{\left(x-1\right)^2}\)
L'Hospital: \(\Rightarrow...=\lim\limits_{x\rightarrow1}\dfrac{2022x^{2021}-2022}{2\left(x-1\right)}=\lim\limits_{x\rightarrow1}\dfrac{2022.2021.x^{2020}}{2}=2043231\)
Is that true :v?
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:\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7\right)^{\dfrac{1}{2}}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{2}\left(2x+7\right)^{-\dfrac{1}{2}}.2+1}{3x^2-8x}=\dfrac{9^{-\dfrac{1}{2}}+1}{3-8}=-\dfrac{4}{15}\)
MG
1
13 tháng 3 2023
\(=\lim\limits_{x->1}\dfrac{\left(x-1\right)\left(x-2\right)}{\sqrt{3x+1}-2-2\left(\sqrt[3]{2x-1}-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x-2\right)}{\dfrac{3x+1-4}{\sqrt{3x+1}+2}-2\cdot\dfrac{2x-1-1}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1}}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-2\right)}{\dfrac{3}{\sqrt{3x+1}+2}-\dfrac{4}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1}}\)
\(=\dfrac{1-2}{\dfrac{3}{\sqrt{3+1}+2}-\dfrac{4}{\sqrt[3]{\left(2\cdot1-1\right)^2}+\sqrt[3]{2\cdot1-1}+1}}\)
\(=-1:\dfrac{-7}{12}=\dfrac{12}{7}\)
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/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+x\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{x\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)