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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:\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7\right)^{\dfrac{1}{2}}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{2}\left(2x+7\right)^{-\dfrac{1}{2}}.2+1}{3x^2-8x}=\dfrac{9^{-\dfrac{1}{2}}+1}{3-8}=-\dfrac{4}{15}\)
Lời giải:
\(L=\lim\limits_{x\to 1}\frac{\sqrt{2x-1}(\sqrt[3]{x+7}-2)+2(\sqrt{2x-1}-1)}{x(x-1)}=\lim\limits_{x\to 1}\frac{\sqrt{2x-1}.\frac{1}{\sqrt[3]{(x+7)^2}+2\sqrt[3]{x+7}+4}+4.\frac{1}{\sqrt{2x-1}+1}}{x}=\frac{25}{12}\)
\(\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x}+\sqrt{x-1}-1}{\sqrt{x^2-1}}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{\dfrac{\left(x-1\right)}{\sqrt{x}+1}+\left(\sqrt{x-1}\right)}{\sqrt{\left(x-1\right)\left(x+1\right)}}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{\left(\sqrt{x-1}\right)\left(\dfrac{\sqrt{x-1}}{\sqrt{x}+1}+1\right)}{\sqrt{x-1}\cdot\sqrt{x+1}}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{\left(\dfrac{\sqrt{x-1}}{\sqrt{x}+1}+1\right)}{\sqrt{x+1}}=\dfrac{\dfrac{\sqrt{1-1}}{\sqrt{1}+1}+1}{\sqrt{1+1}}\)
\(=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
a) Ta có (x - 2)2 = 0 và (x - 2)2 > 0 với ∀x ≠ 2 và
(3x - 5) = 3.2 - 5 = 1 > 0.
Do đó
= +∞.
b) Ta có (x - 1) và x - 1 < 0 với ∀x < 1 và
(2x - 7) = 2.1 - 7 = -5 <0.
Do đó
= +∞.
c) Ta có (x - 1) = 0 và x - 1 > 0 với ∀x > 1 và
(2x - 7) = 2.1 - 7 = -5 < 0.
Do đó
= -∞.
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 \(\left\{{}\begin{matrix}u_1=x\\q=x\end{matrix}\right.\Rightarrow S=u_1.\dfrac{q^{2021}-1}{q-1}=x.\dfrac{x^{2021}-1}{x-1}\)
\(\Rightarrow\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^{2022}-x}{x-1}-2021}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{x^{2022}-x-2021x+2021}{\left(x-1\right)^2}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^{2022}}{x^2}-\dfrac{x}{x^2}-\dfrac{2021x}{x^2}+\dfrac{2021}{x^2}}{\dfrac{x^2}{x^2}-\dfrac{2x}{x^2}+\dfrac{1}{x^2}}=\lim\limits_{x\rightarrow1}\dfrac{x^{2020}}{1}=1\)
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?
\(=\lim\limits_{x\rightarrow1}\dfrac{2x-\dfrac{1}{2}.x^{-\dfrac{1}{2}}}{\dfrac{1}{2}.x^{-\dfrac{1}{2}}}=\dfrac{2-\dfrac{1}{2}}{\dfrac{1}{2}}=3\)
\(=lim_{x\rightarrow1}\dfrac{-x^2+2x-1}{x\left(x-1\right)\left(\sqrt{2x-x^2}+1\right)}=lim_{x\rightarrow1}\dfrac{-\left(x-1\right)^2}{x\left(x-1\right)\left(\sqrt{2x-x^2}+1\right)}\)
\(=lim_{x\rightarrow1}-\dfrac{\left(x-1\right)}{x\left(\sqrt{2x-x^2}+1\right)}\)=0