lim x → + ∞ 3 x - 5 sin 2 x + c o s 2 x x 2 + 2 bằng:
A. -∞.
B. 0.
C. 3.
D. +∞.
K
Khách
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NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 4 2020
\(\lim\limits_{x\rightarrow a}\frac{sin\left(\frac{x-a}{2}\right)}{\frac{x-a}{2}}.cos\left(\frac{x+a}{2}\right)=1.cos\left(\frac{a+a}{2}\right)=cosa\)
b/ \(\lim\limits_{x\rightarrow\pi}\frac{sin\frac{\pi}{2}-sin\frac{x}{2}}{\pi-x}=\lim\limits_{x\rightarrow\pi}\frac{sin\left(\frac{\pi-x}{4}\right)}{\frac{\pi-x}{4}}.\frac{cos\left(\frac{\pi+x}{4}\right)}{2}=\frac{cos\left(\frac{\pi+\pi}{4}\right)}{2}=0\)
c/ Đặt \(x-\frac{\pi}{3}=a\Rightarrow x=a+\frac{\pi}{3}\)
\(\lim\limits_{a\rightarrow0}\frac{sina}{1-2cos\left(a+\frac{\pi}{3}\right)}=\lim\limits_{a\rightarrow0}\frac{sina}{1-cosa+\sqrt{3}sina}\)
\(=\lim\limits_{a\rightarrow0}\frac{2sin\frac{a}{2}cos\frac{a}{2}}{-2sin^2\frac{a}{2}+2\sqrt{3}sin\frac{a}{2}cos\frac{a}{2}}=\lim\limits_{a\rightarrow0}\frac{cos\frac{a}{2}}{-sin\frac{a}{2}+\sqrt{3}cos\frac{a}{2}}=\frac{1}{\sqrt{3}}\)
d/Ta có: \(tana-tanb=\frac{sina}{cosa}-\frac{sinb}{cosb}=\frac{sina.cosb-cosa.sinb}{cosa.cosb}=\frac{sin\left(a-b\right)}{cosa.cosb}\)
Áp dụng:
\(\lim\limits_{x\rightarrow a}\frac{\left(tanx-tana\right)\left(tanx+tana\right)}{\frac{sin\left(x-a\right)}{cos\left(x-a\right)}}=\lim\limits_{x\rightarrow a}\frac{sin\left(x-a\right)\left(tanx+tana\right).cos\left(x-a\right)}{sin\left(x-a\right).cosx.cosa}=\lim\limits_{x\rightarrow a}\frac{\left(tanx+tana\right).cos\left(x-a\right)}{cosx.cosa}\)
\(=\frac{2tana}{cos^2a}\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2020
Bài 1:
\(a=\lim\limits_{x\rightarrow+\infty}\frac{\frac{1}{x}+\frac{2}{\sqrt{x}}-1}{1+\frac{3}{x}}=-1\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{1+\frac{3}{x^2}-\frac{1}{x^3}}{\frac{1}{\sqrt{x}}+\frac{1}{x^2}}=\frac{1}{0}=+\infty\)
\(c=\lim\limits_{x\rightarrow-\infty}\frac{1-2\sqrt{\frac{1}{x^2}-\frac{1}{x}}}{\frac{1}{x}-1}=\frac{1}{-1}=-1\)
Bài 2:
\(a=\lim\limits_{x\rightarrow0}\frac{1-cosx}{1-cos3x}=\lim\limits_{x\rightarrow0}\frac{sinx}{3sin3x}=\lim\limits_{x\rightarrow0}\frac{\frac{sinx}{x}}{9.\frac{sin3x}{3x}}=\frac{1}{9}\)
\(b=\lim\limits_{x\rightarrow0}\frac{cotx-sinx}{x^3}=\frac{\infty}{0}=+\infty\)
\(c=\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}\)
Mà \(\left|sinx\right|\le1\Rightarrow\left|\frac{sinx}{2x}\right|\le\frac{1}{\left|2x\right|}\)
Mà \(\lim\limits_{x\rightarrow\infty}\frac{1}{2\left|x\right|}=0\Rightarrow\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}=0\)
HT
16 tháng 4 2021
1/ \(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7-9\right)\left(2+\sqrt{x+3}\right)}{\left(4-x-3\right)\left(\sqrt{2x+7}+3\right)}=\lim\limits_{x\rightarrow1}\dfrac{2\left(x-1\right)\left(2+\sqrt{x+3}\right)}{\left(x-1\right)\left(-\sqrt{2x+7}-3\right)}=\dfrac{2.4}{-6}=-\dfrac{4}{3}\)
2/ \(=\lim\limits_{x\rightarrow1^-}\dfrac{2.1-3}{1-1}=-\infty\)
3/ \(=\lim\limits_{x\rightarrow2^+}\dfrac{3-x}{x-2}=+\infty\)
4/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{-\dfrac{8x^3}{x^2}+\dfrac{9x^2}{x^2}+\dfrac{x}{x^2}-\dfrac{1}{x^2}}{\dfrac{5x^2}{x^2}+\dfrac{1}{x^2}}=\lim\limits_{x\rightarrow\pm\infty}\dfrac{-8x}{5}=\pm\infty\)
5/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{\dfrac{x^2}{x^2}}+\dfrac{2x}{x}-\dfrac{1}{x}}{\dfrac{2x}{x}+\dfrac{7}{x}}=\dfrac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 2 2020
Tất cả đều ko phải dạng vô định, bạn cứ thay số vào tính thôi:
\(a=\frac{sin\left(\frac{\pi}{4}\right)}{\frac{\pi}{2}}=\frac{\sqrt{2}}{\pi}\)
\(b=\frac{\sqrt[3]{3.4-4}-\sqrt{6-2}}{3}=\frac{0}{3}=0\)
\(c=0.sin\frac{1}{2}=0\)
DT
0
HT
20 tháng 2 2021
a/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-x+1-x^2-x-1}{\sqrt{x^2-x+1}+\sqrt{x^2+x+1}}=\lim\limits_{x\rightarrow+\infty}\dfrac{-\dfrac{2x}{x}}{\sqrt{\dfrac{x^2}{x^2}-\dfrac{x}{x^2}+\dfrac{1}{x^2}}+\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}}=-\dfrac{2}{1+1}=-1\)
b/ \(=\lim\limits_{x\rightarrow2}\dfrac{4x+1-9}{\left(x-2\right)\left(x+2\right)\left(\sqrt{4x+1}+3\right)}=\lim\limits_{x\rightarrow2}\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)\left(\sqrt{4x+1}+3\right)}=\lim\limits_{x\rightarrow2}\dfrac{4}{\left(x+2\right)\left(\sqrt{4x+1}+3\right)}=\dfrac{4}{\left(2+2\right)\left(\sqrt{4.2+1}+3\right)}=\dfrac{1}{6}\)
c/ \(=\lim\limits_{x\rightarrow-2}\dfrac{2x+5-1}{\left(x-2\right)\left(x+2\right)\left(\sqrt{2x+5}+1\right)}=\lim\limits_{x\rightarrow-2}\dfrac{2}{\left(x-2\right)\left(\sqrt{2x+5}+1\right)}=\dfrac{2}{\left(-2-2\right)\left(\sqrt[2]{2.\left(-2\right)+5}+1\right)}=\dfrac{2}{\left(-4\right).2}=-\dfrac{1}{4}\)
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
AM
9 tháng 2 2022
a. \(lim_{x\rightarrow3}\dfrac{x^3-27}{3x^2-5x-2}=\dfrac{3^3-27}{3.3^2-5.3-2}=\dfrac{0}{10}=0\)
b. \(lim_{x\rightarrow2}\dfrac{\sqrt{x+2}-2}{4x^2-3x-2}=\dfrac{\sqrt{2+2}-2}{4.2^2-3.2-2}=\dfrac{0}{8}=0\)
c. \(lim_{x\rightarrow1}\dfrac{1-x^2}{x^2-5x+4}=lim_{x\rightarrow1}\dfrac{\left(1-x\right)\left(x+1\right)}{\left(x-1\right)\left(x-4\right)}=lim_{x\rightarrow1}\dfrac{-\left(x+1\right)}{x-4}=\dfrac{-\left(1+1\right)}{1-4}=\dfrac{2}{3}\)
d. Câu này mình chịu, nhìn đề hơi lạ so với bình thường hehe
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 \(Lim_{x\rightarrow3}\frac{x^4-27x}{2x^2-3x-9}=9\)
Chọn B.
Ta có:
Vậy