Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
3.
\(x-2y+1=0\Leftrightarrow y=\frac{1}{2}x+\frac{1}{2}\)
\(y'=\frac{2}{\left(x+1\right)^2}\Rightarrow\frac{2}{\left(x+1\right)^2}=\frac{1}{2}\)
\(\Rightarrow\left(x+1\right)^2=4\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=1\\x=-3\Rightarrow y=3\end{matrix}\right.\)
Có 2 tiếp tuyến: \(\left[{}\begin{matrix}y=\frac{1}{2}\left(x-1\right)+1\\y=\frac{1}{2}\left(x+3\right)+3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=\frac{1}{2}x+\frac{1}{2}\left(l\right)\\y=\frac{1}{2}x+\frac{9}{2}\end{matrix}\right.\)
4.
\(\lim\limits\frac{\sqrt{2n^2+1}-3n}{n+2}=\lim\limits\frac{\sqrt{2+\frac{1}{n^2}}-3}{1+\frac{2}{n}}=\sqrt{2}-3\)
\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)
5.
\(\lim\limits_{x\rightarrow a}\frac{2\left(x^2-a^2\right)+a\left(a+1\right)-\left(a+1\right)x}{\left(x-a\right)\left(x+a\right)}=\lim\limits_{x\rightarrow a}\frac{\left(x-a\right)\left(2x+2a\right)-\left(a+1\right)\left(x-a\right)}{\left(x-a\right)\left(x+a\right)}\)
\(=\lim\limits_{x\rightarrow a}\frac{\left(x-a\right)\left(2x+a-1\right)}{\left(x-a\right)\left(x+a\right)}=\lim\limits_{x\rightarrow a}\frac{2x+a-1}{x+a}=\frac{3a-1}{2a}\)
1.
\(f'\left(x\right)=-3x^2+6mx-12=3\left(-x^2+2mx-4\right)=3g\left(x\right)\)
Để \(f'\left(x\right)\le0\) \(\forall x\in R\) \(\Leftrightarrow g\left(x\right)\le0;\forall x\in R\)
\(\Leftrightarrow\Delta'=m^2-4\le0\Rightarrow-2\le m\le2\)
\(\Rightarrow m=\left\{-1;0;1;2\right\}\)
2.
\(f'\left(x\right)=\frac{m^2-20}{\left(2x+m\right)^2}\)
Để \(f'\left(x\right)< 0;\forall x\in\left(0;2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2-20< 0\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{20}< m< \sqrt{20}\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow m=\left\{1;2;3;4\right\}\)
Bài 1:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)
\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)
Bài 2:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)
\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)
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 \(\lim\limits_{x\rightarrow\infty}\frac{1}{2\left|x\right|}=0\Rightarrow\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}=0\)
a.
\(\lim\limits_{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}=\lim_{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{(x-1)^3(3x+1)}=\lim\limits _{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{3x+1}.\lim\limits_{x\to 1+}\frac{1}{(x-1)^3}\)
\(=\frac{1}{4}.(+\infty)=+\infty \)
Hoàn toàn tương tự:
\(\lim\limits_{x\to 1-}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}=-\infty \)
Do đó: \(\lim\limits_{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}\neq \lim\limits_{x\to 1-}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}\) nên không tồn tại \(\lim\limits_{x\to 1}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}\)
b.
\(\lim\limits_{x\to 1+}\frac{x^3-3x^2+2}{x^4-4x+3}=\lim\limits_{x\to 1+}\frac{(x-1)(x^2-2x-2)}{(x-1)^2(x^2+2x+3)}=\lim\limits_{x\to 1+}\frac{x^2-2x-2}{(x-1)(x^2+2x+3)}\)
\(=\lim\limits_{x\to 1+}\frac{x^2-2x-2}{x^2+2x+3}.\lim\limits_{x\to 1+}\frac{1}{x-1}=\frac{-1}{2}.(+\infty)=-\infty \)
Tương tự \(\lim\limits_{x\to 1-}\frac{x^3-3x^2+2}{x^4-4x+3}=+\infty \)
Do đó không tồn tại \(\lim\limits_{x\to 1}\frac{x^3-3x^2+2}{x^4-4x+3}\)
c.
\(\lim\limits_{x\to 1}\frac{x^3-2x-1}{x^5-2x-1}=\frac{1^3-2.1-1}{1^5-2.1-1}=1\)
d.
\(\lim\limits_{x\to -1}\frac{(x+2)^2-1}{x^2-1}=\lim\limits_{x\to -1}\frac{(x+2-1)(x+2+1)}{(x-1)(x+1)}=\lim\limits_{x\to -1}\frac{x+3}{x-1}=-1\)
\(a=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}{\left(x-1\right)\left(x^2+x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x+1\right)\left(x^2+1\right)}{x^2+x-1}=\frac{4}{1}=4\)
\(b=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{x^4-x^3+x^2-x+1}{x^2-x+1}=\frac{5}{3}\)
\(c=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)^2}{\left(x^2+1\right)\left(x^2-9\right)}=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x+3\right)}=\frac{0}{60}=0\)
\(d=\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{x^2-2x+1}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4+1}{2x-2}=\lim\limits_{x\rightarrow1}\frac{120x^4-100x^3}{2}=10\)
\(e=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)
\(f=\lim\limits_{x\rightarrow-2}\frac{\left(x+2\right)\left(x-2\right)\left(x^2+4\right)}{\left(x+2\right)x^2}=\lim\limits_{x\rightarrow-2}\frac{\left(x-2\right)\left(x^2+4\right)}{x^2}=-8\)
Hai câu d, e khai triển thì dài quá nên làm biếng sử dụng L'Hopital
a/ Đề không rõ ràng bạn
Từ câu b trở đi, dễ dàng nhận ra tất cả các hàm số đều liên tục trên R
b/ Xét \(f\left(x\right)=x^3+3x^2-1\)
Ta có: \(f\left(-3\right)=-1\) ; \(f\left(-2\right)=3\)
\(\Rightarrow f\left(-3\right).f\left(-2\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm trên \(\left(-3;-2\right)\)
\(f\left(0\right)=-1\Rightarrow f\left(-2\right).f\left(1\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm trên \(\left(-2;0\right)\)
\(f\left(1\right)=3\Rightarrow f\left(0\right).f\left(1\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm trên \(\left(0;1\right)\)
\(\Rightarrow f\left(x\right)\) luôn có 3 nghiệm phân biệt
c/\(f\left(x\right)=m\left(x-1\right)^3\left(m^2-4\right)+x^4-3\)
\(f\left(-2\right)=13\) ; \(f\left(1\right)=-2\)
\(\Rightarrow f\left(-2\right).f\left(1\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm trên \(\left(-2;1\right)\)
\(f\left(2\right)=13\Rightarrow f\left(1\right).f\left(2\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm trên \(\left(1;2\right)\)
\(\Rightarrow f\left(x\right)\) luôn có ít nhất 2 nghiệm
d/ \(f\left(x\right)=5sin3x+x-10\)
\(f\left(0\right)=-10\)
\(f\left(4\pi\right)=4\pi-10\)
\(\Rightarrow f\left(0\right).f\left(4\pi\right)=-10\left(4\pi-10\right)< 0\)
\(\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(0;4\pi\right)\) hay \(f\left(x\right)\) luôn có nghiệm
a) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+2x}-1}{2x}=\lim\limits_{x\rightarrow0}\frac{2x}{2x\left(\sqrt{1+2x}+1\right)}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{1+2x}+1}=\frac{1}{2}\)
b) \(\lim\limits_{x\rightarrow0}\frac{4x}{\sqrt{9+x}-3}=\lim\limits_{x\rightarrow0}\frac{4x\left(\sqrt{9+x}+3\right)}{x}=\lim\limits_{x\rightarrow0}[4\left(\sqrt{9+x}+3\right)=24\)
c) \(\lim\limits_{x\rightarrow2}\frac{\sqrt{x+7}-3}{x-2}=\lim\limits_{x\rightarrow2}\frac{x-2}{\left(x-2\right)\left(\sqrt{x+7}+3\right)}=\lim\limits_{x\rightarrow2}\frac{1}{\sqrt{x+7}+3}=\frac{1}{6}\)
d) \(\lim\limits_{x\rightarrow1}\frac{3x-2-\sqrt{4x^2-x-2}}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\frac{\left(3x-2\right)^2-\left(4x^2-4x-2\right)}{(x^2-3x+2)\left(3x-2+\sqrt{4x^2-x-2}\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(5x-6\right)}{\left(x-1\right)\left(x-2\right)\left(3x-2+\sqrt{4x^2-x-2}\right)}=\frac{1}{2}\\ \\\\ \\ \\ \\ \)
e)\(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\frac{2x+7-\left(x^2-8x+16\right)}{\left(x-1\right)\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x-9\right)}{\left(x-1\right)\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=\lim\limits_{x\rightarrow1}\frac{x-9}{\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=-8\)
f) \(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}-3}{2-\sqrt{x+3}}=\lim\limits_{x\rightarrow1}\frac{(2x-2)\left(2+\sqrt{x+3}\right)}{\left(1-x\right)\left(\sqrt{2x+7}+3\right)}=\lim\limits_{x\rightarrow1}\frac{-2\left(2+\sqrt{x+3}\right)}{\sqrt{2x+7}+3}=\frac{-4}{3}\)
g) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}=\lim\limits_{x\rightarrow0}\frac{x^2\left(\sqrt{x^2+16}+4\right)}{x^2\left(\sqrt{x^2+1}+1\right)}=4\)
h)
\(\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-\sqrt{2x+1}}{x-4}=\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-3}{x-4}+\lim\limits_{x\rightarrow4}\frac{3-\sqrt{2x+1}}{x-4}=\lim\limits_{x\rightarrow4}\frac{1}{\sqrt{x+5}+4}+\lim\limits_{x\rightarrow4}\frac{8-2x}{\left(x-4\right)\left(3+\sqrt{2x+1}\right)}=\frac{1}{7}-\frac{1}{3}=\frac{-4}{21}\)
k) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}+\sqrt{x+4}-3}{x}=\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}-1}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{x+4}-2}{x}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+1}+1}+\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+4}+2}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)
Vậy nó ko phải dạng vô định, cứ thay số trực tiếp
\(=\frac{2}{0}=+\infty\)
Nếu là mũ 3 thì nó là dạng 0/0 rút gọn được. Nên chắc là đề ghi nhầm đấy
