Tính giới hạn L = lim x → − 1 x 2 − x − 2 3 x 2 + 8 x + 5 .
A. L = 0
B. L = − ∞
C. L = − 3 2 .
D. L = 1 2 .
K
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NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 1
\(=\lim\limits_{x\rightarrow0}\dfrac{2\left(\sqrt[]{2x+1}-1\right)+2-\sqrt[3]{x^2+x+8}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{2.2x}{\sqrt[]{2x+1}+1}-\dfrac{x\left(x+1\right)}{\sqrt[3]{\left(x^2+x+8\right)^2}+2\sqrt[3]{x^2+x+8}+4}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{2x+1}+1}-\dfrac{x+1}{\sqrt[3]{\left(x^2+x+8\right)^2}+2\sqrt[3]{x^2+x+8}+4}\right)\)
\(=\dfrac{23}{12}\)
AH
Akai Haruma
Giáo viên
3 tháng 4 2022
Lời giải:
\(L=\lim\limits_{x\to 1}\frac{\sqrt{2x-1}(\sqrt[3]{x+7}-2)+2(\sqrt{2x-1}-1)}{x(x-1)}=\lim\limits_{x\to 1}\frac{\sqrt{2x-1}.\frac{1}{\sqrt[3]{(x+7)^2}+2\sqrt[3]{x+7}+4}+4.\frac{1}{\sqrt{2x-1}+1}}{x}=\frac{25}{12}\)
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 \(3-x>0;\forall x< 3\)
\(\Rightarrow\lim\limits_{x\rightarrow3^-}\dfrac{-x-3}{3-x}=-\infty\)
MH
3 tháng 2 2021
a) \(\lim\limits_{x\rightarrow-2}\dfrac{2x^2+x-6}{x^3+8}=\lim\limits_{x\rightarrow-2}\dfrac{\left(2x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\\ =\lim\limits_{x\rightarrow-2}\dfrac{2x-3}{x^2-2x+4}=-\dfrac{7}{12}\).
b) \(\lim\limits_{x\rightarrow3}\dfrac{x^4-x^2-72}{x^2-2x-3}=\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}\\ =\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)}{x+1}=\dfrac{51}{2}\).
c) \(\lim\limits_{x\rightarrow-1}\dfrac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\lim\limits_{x\rightarrow-1}\dfrac{x^4-x^3+x^2-x+1}{x^2-x+1}=\dfrac{5}{3}\).
d) \(\lim\limits_{x\rightarrow1}\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right)\\ =\lim\limits_{x\rightarrow1}\dfrac{1-x}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{-1}{x+1}=-\dfrac{1}{2}\).
MA
2
KB
2 tháng 5 2022
Ta có : (...) = \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x^3}-\left(x+1\right)-\left[\sqrt[3]{x^2+7}-\left(x+1\right)\right]}{x^2-1}\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x^3}-\left(x+1\right)}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{5-x^3-\left(x+1\right)^2}{\left(\sqrt{5-x^3}+x+1\right)\left(x^2-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{-x^3-x^2-2x+4}{...}\) \(=\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+2x+4\right)\left(x-1\right)}{...}\)
= \(\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+2x+4\right)}{\left(x+1\right)\left(\sqrt{5-x^3}+x+1\right)}=\dfrac{-7}{8}\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-\left(x+1\right)}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{x^2+7-x^3-3x^2-3x-1}{\left(x^2-1\right)\left[\sqrt[3]{\left(x+7\right)^2}+\left(x+1\right)\sqrt[3]{x^2+7}+\left(x+1\right)^2\right]}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+3x+6\right)\left(x-1\right)}{...}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+3x+6\right)}{\left(x+1\right)\left[\sqrt[3]{\left(x^2+7\right)^2}+\sqrt[3]{x^2+7}\left(x+1\right)+\left(x+1\right)^2\right]}\)
\(=\dfrac{-\left(1+3+6\right)}{\left(1+1\right)\left(4+2.2+4\right)}=\dfrac{-5}{12}\)
Suy ra : \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}=\dfrac{-7}{8}+\dfrac{5}{12}=\dfrac{-11}{24}\)
10 tháng 11 2023
a: \(\lim\limits_{x\rightarrow3}\dfrac{\sqrt{2x+10}-4}{3x-9}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{2x+10-16}{3x-9}\cdot\dfrac{1}{\sqrt{2x+10}+4}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{2\left(x-3\right)}{3\left(x-3\right)\cdot\left(\sqrt{2x+10}+4\right)}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{2}{3\left(\sqrt{2x+10}+4\right)}\)
\(=\dfrac{2}{3\cdot\sqrt{6+10}+3\cdot4}=\dfrac{2}{3\cdot4+3\cdot4}=\dfrac{2}{24}=\dfrac{1}{12}\)
b: \(\lim\limits_{x\rightarrow7}\dfrac{\sqrt{4x+8}-6}{x^2-9x+14}\)
\(=\lim\limits_{x\rightarrow7}\dfrac{4x+8-36}{\sqrt{4x+8}+6}\cdot\dfrac{1}{\left(x-2\right)\left(x-7\right)}\)
\(=\lim\limits_{x\rightarrow7}\dfrac{4x-28}{\left(\sqrt{4x+8}+6\right)\cdot\left(x-2\right)\left(x-7\right)}\)
\(=\lim\limits_{x\rightarrow7}\dfrac{4}{\left(\sqrt{4x+8}+6\right)\left(x-2\right)}\)
\(=\dfrac{4}{\left(\sqrt{4\cdot7+8}+6\right)\left(7-2\right)}\)
\(=\dfrac{4}{5\cdot12}=\dfrac{4}{60}=\dfrac{1}{15}\)
c: \(\lim\limits_{x\rightarrow5}\dfrac{x^2-8x+15}{2x^2-9x-5}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\left(x-3\right)\left(x-5\right)}{2x^2-10x+x-5}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\left(x-3\right)\left(x-5\right)}{\left(x-5\right)\left(2x+1\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{x-3}{2x+1}=\dfrac{5-3}{2\cdot5+1}=\dfrac{2}{11}\)
HT
9 tháng 2 2021
a/ \(\lim\limits_{x\rightarrow2}\dfrac{2+3}{4+2+4}=\dfrac{5}{10}=\dfrac{1}{2}\)
b/ \(\lim\limits_{x\rightarrow-3}\dfrac{\left(x+2\right)\left(x+3\right)}{x\left(x+3\right)}=\lim\limits_{x\rightarrow-3}\dfrac{x+2}{x}=\dfrac{-3+2}{-3}=\dfrac{1}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 3 2021
a. Chắc đề là: \(\lim\dfrac{2-5^{n-2}}{3^n+2.5^n}=\lim\dfrac{2\left(\dfrac{1}{5}\right)^{n-2}-1}{9\left(\dfrac{3}{5}\right)^{n-2}+50}=-\dfrac{1}{50}\)
b. \(=\lim\dfrac{2\left(\dfrac{1}{5}\right)^n-25}{\left(\dfrac{3}{5}\right)^n-2}=\dfrac{25}{2}\)
2.
Đặt \(f\left(x\right)=x^4+x^3-3x^2+x+1\)
Hàm f(x) liên tục trên R
\(f\left(0\right)=1>0\) ; \(f\left(-1\right)=-3< 0\)
\(\Rightarrow f\left(0\right).f\left(-1\right)< 0\Rightarrow f\left(x\right)=0\) luôn có ít nhất 1 nghiệm thuộc khoảng \(\left(-1;0\right)\)
Hay pt đã cho luôn có ít nhất 1 nghiệm âm lớn hơn -1
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 3 2021
3.
Ta có: M là trung điểm AD, N là trung điểm SD
\(\Rightarrow\) MN là đường trung bình tam giác SAD
\(\Rightarrow MN||SA\Rightarrow\left(MN,SC\right)=\left(SA,SC\right)\)
Ta có: \(AC=\sqrt{AB^2+BC^2}=a\sqrt{2}\)
\(SA=SC=a\)
\(\Rightarrow SA^2+SC^2=AC^2\Rightarrow\Delta SAC\) vuông tại S hay \(SA\perp SC\)
\(\Rightarrow\) Góc giữa MN và SC bằng 90 độ
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: \(=lim_{x->+\infty}x\sqrt{\dfrac{1+\dfrac{1}{x}+\dfrac{3}{x^2}}{3x^2+4-\dfrac{5}{x^2}}}\)
\(=lim_{x->+\infty}x\sqrt{\dfrac{1}{3x^2+4}}=+\infty\)
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