Tính giới hạn y = lim x → 1 ( x 2 - 4 x + 7 x + 1 )
A. I = 4
B. I = 5
C. I = -4
D. I = 2
K
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CM
Cao Minh Tâm
23 tháng 1 2018
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NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 3 2022
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-\left(x+1\right)}{2x^2-x}=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{x^2+1}-\left(x+1\right)\right)\left(\sqrt{x^2+1}+x+1\right)}{x\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{-2x}{x\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{-2}{\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)
\(=\dfrac{-2}{\left(0-1\right)\left(\sqrt{1}+1\right)}=1\)
a. \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\lim\limits_{x\rightarrow2}\dfrac{1}{x+2}=\dfrac{1}{4}\)
b. \(\lim\limits_{x\rightarrow3^-}\dfrac{x+3}{x-3}=\lim\limits_{x\rightarrow3^-}\dfrac{-x-3}{3-x}\)
Do \(\lim\limits_{x\rightarrow3^-}\left(-x-3\right)=-6< 0\)
\(\lim\limits_{x\rightarrow3^-}\left(3-x\right)=0\) và \(3-x>0;\forall x< 3\)
\(\Rightarrow\lim\limits_{x\rightarrow3^-}\dfrac{-x-3}{3-x}=-\infty\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2020
\(a=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}{\left(x-1\right)\left(x^2+x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x+1\right)\left(x^2+1\right)}{x^2+x-1}=\frac{4}{1}=4\)
\(b=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{x^4-x^3+x^2-x+1}{x^2-x+1}=\frac{5}{3}\)
\(c=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)^2}{\left(x^2+1\right)\left(x^2-9\right)}=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x+3\right)}=\frac{0}{60}=0\)
\(d=\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{x^2-2x+1}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4+1}{2x-2}=\lim\limits_{x\rightarrow1}\frac{120x^4-100x^3}{2}=10\)
\(e=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)
\(f=\lim\limits_{x\rightarrow-2}\frac{\left(x+2\right)\left(x-2\right)\left(x^2+4\right)}{\left(x+2\right)x^2}=\lim\limits_{x\rightarrow-2}\frac{\left(x-2\right)\left(x^2+4\right)}{x^2}=-8\)
Hai câu d, e khai triển thì dài quá nên làm biếng sử dụng L'Hopital
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 2 2020
\(a=\lim\limits_{x\rightarrow3}\frac{\left(x-3\right)\left(2x+3\right)}{\left(x-3\right)\left(x^3+3x^2+9x\right)}=\lim\limits_{x\rightarrow3}\frac{2x+3}{x^3+3x^2+9x}=\frac{2.3+3}{3^3+2.3^2+9.3}=...\)
\(b=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x^4+x^2+2x^3+2x+2\right)}=\frac{1+1}{1+1+2+2+2}=...\)
\(c=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)^2\left(4x^3+3x^2+2x+1\right)}{\left(x-1\right)^2\left(x^2+x+2\right)}=\frac{4+3+2+1}{1+1+2}=...\)
\(d=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{1+1+1+1+1}{1+1+1}=...\)
PN
26 tháng 5 2021
\(Lim_{x\rightarrow3}\frac{x^4-27x}{2x^2-3x-9}=Lim_{x\rightarrow3}\frac{x\left(x^3-3^3\right)}{\left(x-3\right)\left(2x+3\right)}\)
\(=Lim_{x\rightarrow3}\frac{x\left(x-3\right)\left(x^2+3x+9\right)}{\left(x-3\right)\left(2x+3\right)}=Lim_{x\rightarrow3}\frac{x\left(x^2+3x+9\right)}{2x+3}\)
\(=\frac{3\left(3^2+3.3+9\right)}{3.2+3}=\frac{3\left(9+9+9\right)}{9}=9\)
Vậy \(Lim_{x\rightarrow3}\frac{x^4-27x}{2x^2-3x-9}=9\)
19 tháng 2 2023
a: \(\lim\limits_{x->0^-^-}\dfrac{-2x+x}{x\left(x-1\right)}=lim_{x->0^-}\left(\dfrac{-x}{x\left(x-1\right)}\right)\)
\(=lim_{x->0^-}\left(\dfrac{-1}{x-1}\right)=\dfrac{-1}{0-1}=\dfrac{-1}{-1}=1\)
b: \(=lim_{x->-\infty}\left(\dfrac{x^2-x-x^2+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)
\(=lim_{x->-\infty}\left(\dfrac{-x+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)
\(=lim_{x->-\infty}\left(\dfrac{-1+\dfrac{1}{x}}{-\sqrt{1-\dfrac{1}{x^2}}-\sqrt{1-\dfrac{1}{x^2}}}\right)=\dfrac{-1}{-2}=\dfrac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 2 2020
a/ Do \(x\rightarrow-3^+\) nên \(x>-3\Rightarrow x+3>0\Rightarrow\left|x+3\right|=x+3\)
\(\Rightarrow\lim\limits_{x\rightarrow-3^+}\frac{3x+9}{\left|x+3\right|}=\lim\limits_{x\rightarrow-3^+}\frac{3\left(x+3\right)}{x+3}=3\)
b/ \(=\lim\limits_{x\rightarrow0^+}\frac{\sqrt{x}\left(1-3\sqrt{x}\right)}{\sqrt{x}\left(4\sqrt{x}-2\right)}=\lim\limits_{x\rightarrow0^+}\frac{1-3\sqrt{x}}{4\sqrt{x}-2}=-\frac{1}{2}\)
Ở câu này \(x\rightarrow0^+\) có nghĩa \(x>0\), nó chỉ để căn thức xác định, ngoài ra ko có gì đặc biệt hết
c/ Tương tự câu c, cũng chỉ để căn thức xác định \(\left(x< 1\right)\)
\(\lim\limits_{x\rightarrow1^-}\frac{\sqrt{1-x}}{\left(1-x\right)\left(x+4\right)}=\lim\limits_{x\rightarrow1^-}\frac{1}{\sqrt{1-x}\left(x+4\right)}=+\infty\)
d/ Chắc bạn ghi nhầm đề, đây ko phải giới hạn dạng vô định (vì tử khác 0, mẫu bằng 0):
\(x\rightarrow\sqrt{2}^-\Rightarrow x< \sqrt{2}\Rightarrow x^4-4< 0\)
\(\Rightarrow\lim\limits_{x\rightarrow\sqrt{2}^-}\frac{\left|x-2\right|}{x^4-4}=-\infty\)
30 tháng 1 2023
a: \(=lim_{x->-\infty}\dfrac{2x-5+\dfrac{1}{x^2}}{7-\dfrac{1}{x}+\dfrac{4}{x^2}}\)
\(=\dfrac{2x-5}{7}\)
\(=\dfrac{2}{7}x-\dfrac{5}{7}\)
\(=-\infty\)
b: \(=lim_{x->+\infty}x\sqrt{\dfrac{1+\dfrac{1}{x}+\dfrac{3}{x^2}}{3x^2+4-\dfrac{5}{x^2}}}\)
\(=lim_{x->+\infty}x\sqrt{\dfrac{1}{3x^2+4}}=+\infty\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 3 2020
\(A=\lim\limits_{x\rightarrow2}\frac{\left(x-2\right)\left(2x-1\right)}{x-2}=\lim\limits_{x\rightarrow2}\left(2x-1\right)=3\)
\(B=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x^2-2x+3\right)}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\frac{x^2-2x+3}{x+1}=\frac{1-2+3}{1+1}=1\)
\(C=\lim\limits_{x\rightarrow2}\frac{x^2+2x}{x^2+4x+4}=\frac{4+4}{4+8+4}=\frac{1}{2}\)
\(D=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x^2-1\right)}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow1}\frac{x^2-1}{x-2}=\frac{0}{-1}=0\)
\(E=\lim\limits_{x\rightarrow1}\frac{x^3-5x^2+3x+9}{x^4-8x^4-9}=\frac{1-5+3+9}{1-8-9}=-\frac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 3 2020
\(F=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x-1\right)\left(x^2+1\right)}{\left(x+1\right)\left(x^2-3x+3\right)}=\lim\limits_{x\rightarrow-1}\frac{\left(x-1\right)\left(x^2+1\right)}{x^2-3x+3}=\frac{-2.2}{1+3+3}=-\frac{2}{5}\)
\(G=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(2x+1\right)}=\lim\limits_{x\rightarrow1}\frac{x+3}{2x+1}=\frac{4}{3}\)
\(H=\lim\limits_{x\rightarrow-2}\frac{\left(x+2\right)\left(x-1\right)^2}{\left(2-x\right)\left(x+2\right)}=\lim\limits_{x\rightarrow-2}\frac{\left(x-1\right)^2}{2-x}=\frac{9}{4}\)
\(I=\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+1}{x^2-1}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4}{2x}=\frac{24-25}{2}=-\frac{1}{2}\)
\(K=\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)