lim 4x+3 / x-1
x--> 2
Những câu hỏi liên quan
Giới hạn
lim
x
→
3
x
+
1
-
5
x
+
1
x
-
4
x
-
3...
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Giới hạn lim x → 3 x + 1 - 5 x + 1 x - 4 x - 3 bằng a b (phân số tối giản). Giá trị của a - b là:
A. 1
B. 1 9
C. -1
D. 2
11) \(\lim\limits_{x->1}\) \(\dfrac{3_{\sqrt{4x-1}-\sqrt{4x-3}}}{x-1}\)
11) \(\lim\limits_{x->4}\dfrac{4x-1}{x^2-8x+16}\)
12) \(\lim\limits_{x->2}\)\(\dfrac{4-x^2}{x^3-8}\)
13) \(\lim\limits_{x->+\infty}\left(3_{\sqrt{x^3+4x^2}-x}\right)\)
Giúp mình với ạ
1) lim\(\dfrac{x-5x^2+1}{x^2-1}\)(x-->-∞)
2) lim\(\dfrac{5x^3\left(2-x^2\right)^3\left(4x^2+1\right)^2}{4x^{13}+x^2-6}\)(x-->+∞)
3) lim\(\dfrac{4x-\sqrt{9x^2+x}}{3-x}\)(x-->+∞)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{x-5x^2+1}{x^2-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}-5+\dfrac{1}{x^2}}{1-\dfrac{1}{x^2}}=\dfrac{-5}{1}=-5\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{5x^3\left(2-x^2\right)^3\left(4x^2+1\right)^2}{4x^{13}+x^2-6}=\lim\limits_{x\rightarrow+\infty}\dfrac{5\left(\dfrac{2}{x^2}-1\right)^3\left(4+\dfrac{1}{x^2}\right)^2}{4+\dfrac{1}{x^{11}}-\dfrac{6}{x^{13}}}=\dfrac{5.\left(-1\right)^3.4^2}{4}=-20\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{4x-\sqrt{9x^2+x}}{3-x}=\lim\limits_{x\rightarrow+\infty}\dfrac{4-\sqrt{9+\dfrac{1}{x}}}{\dfrac{3}{x}-1}=\dfrac{4-3}{-1}=-1\)
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\(\lim\limits_{x\rightarrow1}\dfrac{x^3-3x^2+2}{x^2-4x+3}\)
\(\lim\limits_{x\rightarrow1^-}\dfrac{x^2+3x+2}{\left|x+1\right|}\)
\(\lim\limits_{x\rightarrow3}\dfrac{\sqrt[3]{x+5}-2}{x^2-4x+3}\)
\(a=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x^2-2x-2\right)}{\left(x-1\right)\left(x-3\right)}=\lim\limits_{x\rightarrow1}\dfrac{x^2-2x-2}{x-3}=\dfrac{3}{2}\)
Câu b bạn coi lại đề, là \(x\rightarrow-1^-\) hay \(x\rightarrow1^-\) (đúng như đề thì ko phải dạng vô định, cứ thay số rồi bấm máy)
\(c=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)}{\left(x-3\right)\left(x-1\right)\left(\sqrt[3]{\left(x+5\right)^2}+2\sqrt[3]{x+5}+4\right)}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{1}{\left(x-1\right)\left(\sqrt[3]{\left(x+5\right)^2}+2\sqrt[3]{x+5}+4\right)}=\dfrac{1}{2.\left(4+4+4\right)}=...\)
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a/ \(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}{\left(x-1\right)\left(x-3\right)}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}{x-3}=....\)
Từ 2 câu kia lát tui làm, ăn cơm đã :D
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Tìm giới hạn1) xrightarrow[x-3]{lim}dfrac{x^2-5x+6}{sqrt{2x+3}-sqrt{4x-3}} 2) xrightarrow[x-1]{lim}dfrac{sqrt{x^2+2}-sqrt{4x-1}}{x-1} 3) xrightarrow[x--1]{lim}dfrac{x-2}{xleft|x+1right|} 4) xrightarrow[x-a]{lim}dfrac{x^n-a^n}{x-a}5) xrightarrow[x-1]{lim}(dfrac{n}{1-x^n}-dfrac{1}{1-x})6) xrightarrow[x-1]{lim}dfrac{x^n-nx+n-1}{left(x-1right)^2}
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Tìm giới hạn
1) \(\xrightarrow[x->3]{lim}\dfrac{x^2-5x+6}{\sqrt{2x+3}-\sqrt{4x-3}}\)
2) \(\xrightarrow[x->1]{lim}\dfrac{\sqrt{x^2+2}-\sqrt{4x-1}}{x-1}\)
3) \(\xrightarrow[x->-1]{lim}\dfrac{x-2}{x\left|x+1\right|}\)
4) \(\xrightarrow[x->a]{lim}\dfrac{x^n-a^n}{x-a}\)
5) \(\xrightarrow[x->1]{lim}(\dfrac{n}{1-x^n}-\dfrac{1}{1-x})\)
6) \(\xrightarrow[x->1]{lim}\dfrac{x^n-nx+n-1}{\left(x-1\right)^2}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-\sqrt[3]{2x+1}}{x}\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{4x+5}-3}{\sqrt[3]{5x+3}-2}\)
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt[4]{2x+3}+\sqrt[3]{2+3x}}{\sqrt{x+2}-1}\)
\(a=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-1+1-\sqrt[3]{2x+1}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x}{\sqrt[]{4x+1}+1}+\dfrac{-2x}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{4x+1}+1}+\dfrac{-2}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}\right)=...\)
\(b=\lim\limits_{x\rightarrow1}\dfrac{4\left(x-1\right)\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(x-1\right)\left(\sqrt[]{4x+5}+3\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{4\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(\sqrt[]{4x+5}+3\right)}=...\)
\(c=\lim\limits_{x\rightarrow-1}\dfrac{\left(2x+3\right)^{\dfrac{1}{4}}+\left(2+3x\right)^{\dfrac{1}{3}}}{\left(x+2\right)^{\dfrac{1}{2}}-1}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{1}{2}\left(2x+3\right)^{-\dfrac{3}{4}}+\left(2+3x\right)^{-\dfrac{2}{3}}}{\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}=3\)
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\(\lim\limits_{x\rightarrow3}\dfrac{\sqrt{2x+3}-x}{x^2-4x+3}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{x+1}-1}{\sqrt[4]{2x+1}-1}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^2}\)
\(a=\lim\limits_{x\rightarrow3}\dfrac{2x+3-x^2}{\left(x^2-4x+3\right)\left(\sqrt[]{2x+3}+x\right)}=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)\left(-x-1\right)}{\left(x-3\right)\left(x-1\right)\left(\sqrt[]{2x+3}+x\right)}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{-x-1}{\left(x-1\right)\left(\sqrt[]{2x+3}+x\right)}=...\)
\(b=\lim\limits_{x\rightarrow0}\dfrac{\left(x+1\right)^{\dfrac{1}{3}}-1}{\left(2x+1\right)^{\dfrac{1}{4}}-1}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{3}\left(x+1\right)^{-\dfrac{2}{3}}}{\dfrac{1}{2}\left(2x+1\right)^{-\dfrac{3}{4}}}=\dfrac{2}{3}\)
\(c=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt[]{1+4x}-2x-1\right)+\left(2x+1-\sqrt[3]{1+6x}\right)}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{-4x^2}{2x+1+\sqrt[]{4x+1}}+\dfrac{x^2\left(8x+12\right)}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{-4}{2x+1+\sqrt[]{4x+1}}+\dfrac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}\right)=...\)
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\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2-x+1}-x\right)\)
\(\lim\limits_{x\rightarrow-\infty}x\left(\sqrt{4x^2+1}-x\right)\)
\(\lim\limits_{x\rightarrow-\infty}\left(4x^5-3x^3+x+1\right)\)
\(\lim\limits_{x\rightarrow+\infty}\sqrt{x^4-x^3+x^2-x}\)
Hic nan qua :( Lam vay
P/s: Anh Lam check all ho em nhung bai em lam nhe :( Em cam on
1/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-x+1-x^2}{\sqrt{x^2-x+1}+x}=\dfrac{-1}{1+1}=-\dfrac{1}{2}\)
2/ \(=\lim\limits_{x\rightarrow-\infty}x\left(\dfrac{4x^2+1-x^2}{\sqrt{4x^2+1}+x}\right)=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x}{x}}{-\sqrt{\dfrac{4x^2}{x^2}+\dfrac{1}{x^2}}+\dfrac{x}{x}}=\dfrac{1}{-2+1}=-1\)
3/ \(=\lim\limits_{x\rightarrow-\infty}x^5\left(4-\dfrac{3}{x^2}+\dfrac{1}{x^4}+\dfrac{1}{x^5}\right)=-\infty\)
4/ \(=\lim\limits_{x\rightarrow+\infty}\sqrt{x^4}\left(\sqrt{1-\dfrac{x^3}{x^4}+\dfrac{x^2}{x^4}-\dfrac{x}{x^4}}\right)=+\infty\)
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Cho các phát biểu sau (1): Hàm số
y
x
3
+
3
x
2
+
3
x
+
1
có đồ thị là (C) không có cực trị (2).Hàm số
y
x
3
+
3
x
2
+
3
x
+
1
có điểm uốn là U(-1;0) (3). Đồ thị hàm số
y
3
x
-
2...
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Cho các phát biểu sau (1): Hàm số y = x 3 + 3 x 2 + 3 x + 1 có đồ thị là (C) không có cực trị (2).Hàm số y = x 3 + 3 x 2 + 3 x + 1 có điểm uốn là U(-1;0) (3). Đồ thị hàm số y = 3 x - 2 x - 2 có dạng.Hàm số y = 2 x + 1 x + 1 có lim x → 1 + 2 x + 1 x + 1 = - ∞ và lim x → 1 - 2 x + 1 x + 1 = + ∞ .Số các phát biểu đúng là:
A. 1
B. 2
C. 3
D. 0
