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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:\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7\right)^{\dfrac{1}{2}}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{2}\left(2x+7\right)^{-\dfrac{1}{2}}.2+1}{3x^2-8x}=\dfrac{9^{-\dfrac{1}{2}}+1}{3-8}=-\dfrac{4}{15}\)
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
câu 1 bạn lm kiểu j vậy chả hiểu luôn bạn có thể lm lại chi tiết hơn dc ko
Lời giải:
a. \(\lim\limits_{x\to 1+}(x^3+x+1)=3>0\)
\(\lim\limits_{x\to 1+}(x-1)=0\) và $x-1>0$ khi $x>1$
\(\Rightarrow \lim\limits_{x\to 1+}\frac{x^3+x+1}{x-1}=+\infty\)
b.
\(\lim\limits_{x\to -1+}(3x+2)=-1<0\)
\(\lim\limits_{x\to -1+}(x+1)=0\) và $x+1>0$ khi $x>-1$
\(\Rightarrow \lim\limits_{x\to -1+}\frac{3x+2}{x+1}=-\infty\)
c.
\(\lim\limits_{x\to 2-}(x-15)=-17<0\)
\(\lim\limits_{x\to 2-}(x-2)=0\) và $x-2<0$ khi $x<2$
\(\Rightarrow \lim\limits_{x\to 2-}\frac{x-15}{x-2}=+\infty\)
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: \(\lim\limits_{x\rightarrow2}\dfrac{x^2-2x}{\sqrt{x+2}-2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)}{\dfrac{x+2-4}{\sqrt{x+2}+2}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)\cdot\left(\sqrt{x+2}+2\right)}{x-2}\)
\(=\lim\limits_{x\rightarrow2}x\left(\sqrt{x+2}+2\right)\)
\(=2\cdot\left(\sqrt{2+2}+2\right)\)
\(=2\cdot4=8\)
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?
\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-2x+1}{x-1}=3\rightarrow\lim\limits_{x\rightarrow1}\left(f\left(x\right)-2x+1\right)=0\\ \rightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=1\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{3f\left(x\right)+1}-x-1}{\sqrt{4x+5}-3x-2}=\dfrac{\sqrt{3.1+1}-1-1}{\sqrt{4.1+5}-3.1-2}=0\)
1: \(A=\dfrac{x^2-\left(a+1\right)x+a}{x^3-a^3}\)
\(=\dfrac{x^2-xa-x+a}{\left(x-a\right)\left(x^2+ax+a^2\right)}\)
\(=\dfrac{\left(x-a\right)\left(x-1\right)}{\left(x-a\right)\left(x^2+ax+a^2\right)}=\dfrac{x-1}{x^2+ax+a^2}\)
\(lim_{x->a}A=lim_{x->a}\left(\dfrac{x-1}{x^2+ax+a^2}\right)\)
\(=\dfrac{a-1}{a^2+a^2+a^2}=\dfrac{a-1}{3a^2}\)
2: \(B=\dfrac{1}{1-x}-\dfrac{3}{1-x^3}\)
\(=\dfrac{-1}{x-1}+\dfrac{3}{x^3-1}\)
\(=\dfrac{-x^2-x-1+3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-x^2-x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-x-2}{x^2+x+1}\)
\(lim_{x->1}\left(B\right)=\dfrac{-1-2}{1^2+1+1}=\dfrac{-3}{3}=-1\)
3: \(C=\dfrac{\left(x+h\right)^3-x^3}{h}=\dfrac{\left(x+h-x\right)\left(x^2+2xh+h^2+x^2+hx+x^2\right)}{h}\)
\(=3x^2+3hx\)
\(lim_{h->0}\left(C\right)=3x^2+3\cdot0\cdot x=3x^2\)
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
= -∞.
Giỏi quá ta, chắc là hs cao tuổi nhất ...