tính giới hạn sau:
\(\lim\limits_{x\rightarrow\dfrac{\pi}{6}}\dfrac{\sqrt{3}sinx-cosx}{sin\left(\dfrac{\pi}{3}-2x\right)}\)
Tính giới hạn:
\(_{\lim\limits_{x\rightarrow\dfrac{\Pi}{4}}}\dfrac{\sqrt{2}Cosx-1}{\sqrt{2}Sinx-1}\)
Đặt \(t=x-\dfrac{\pi}{4}\), khi đó:
\(\lim\limits_{x\rightarrow\dfrac{\pi}{4}}\dfrac{\sqrt{2}cosx-1}{\sqrt{2}sinx-1}=\lim\limits_{t\rightarrow0}\dfrac{\sqrt{2}cos\left(t+\dfrac{\pi}{4}\right)-1}{\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)-1}\)
\(=\lim\limits_{t\rightarrow0}\dfrac{cost-sint-1}{cost+sint-1}\)
\(=\lim\limits_{t\rightarrow0}\dfrac{1-2sin^2\dfrac{t}{2}-2sin\dfrac{t}{2}.cos\dfrac{t}{2}-1}{1-2sin^2\dfrac{t}{2}+2sin\dfrac{t}{2}.cos\dfrac{t}{2}-1}\)
\(=\lim\limits_{t\rightarrow0}\dfrac{-2sin\dfrac{t}{2}\left(sin\dfrac{t}{2}+cos\dfrac{t}{2}\right)}{-2sin\dfrac{t}{2}\left(sin\dfrac{t}{2}-cos\dfrac{t}{2}\right)}\)
\(=\lim\limits_{t\rightarrow0}\dfrac{sin\dfrac{t}{2}+cos\dfrac{t}{2}}{sin\dfrac{t}{2}-cos\dfrac{t}{2}}\)
\(=-1\)
lim đề bài cho = \(\lim\limits_{x->\dfrac{\pi}{4}}\dfrac{-\sqrt{2}sinx}{\sqrt{2}cosx}\)
Thay x vào là xong
Tính các giới hạn
a) \(\lim\limits_{x\rightarrow a}\dfrac{\sin x-\sin a}{x-a}\)
b) \(\lim\limits_{x\rightarrow1}\left(1-x\right)\tan\dfrac{\pi x}{2}\)
c) \(\lim\limits_{x\rightarrow\dfrac{\pi}{3}}\dfrac{2\sin^2x+\sin x-1}{2\sin^2x-3\sin x+1}\)
d) \(\lim\limits_{x\rightarrow0}\dfrac{\tan x-\sin x}{\sin^3x}\)
giải phương trình
a) \(sinx=sin\dfrac{\pi}{4}\)
b) \(cos2x=cosx\)
c) \(tan\left(x-\dfrac{\pi}{3}\right)=\sqrt{3}\)
d) \(cot\left(2x+\dfrac{\pi}{6}\right)=cot\dfrac{\pi}{4}\)
a: \(sinx=sin\left(\dfrac{\Omega}{4}\right)\)
=>\(\left[{}\begin{matrix}x=\dfrac{\Omega}{4}+k2\Omega\\x=\Omega-\dfrac{\Omega}{4}+k2\Omega=\dfrac{3}{4}\Omega+k2\Omega\end{matrix}\right.\)
b: cos2x=cosx
=>\(\left[{}\begin{matrix}2x=x+k2\Omega\\2x=-x+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k2\Omega\\3x=k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=k2\Omega\\x=\dfrac{k2\Omega}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{k2\Omega}{3}\)
c:
ĐKXĐ: \(x-\dfrac{\Omega}{3}< >\dfrac{\Omega}{2}+k\Omega\)
=>\(x< >\dfrac{5}{6}\Omega+k\Omega\)
\(tan\left(x-\dfrac{\Omega}{3}\right)=\sqrt{3}\)
=>\(x-\dfrac{\Omega}{3}=\dfrac{\Omega}{3}+k\Omega\)
=>\(x=\dfrac{2}{3}\Omega+k\Omega\)
d:
ĐKXĐ: \(2x+\dfrac{\Omega}{6}< >k\Omega\)
=>\(2x< >-\dfrac{\Omega}{6}+k\Omega\)
=>\(x< >-\dfrac{1}{12}\Omega+\dfrac{k\Omega}{2}\)
\(cot\left(2x+\dfrac{\Omega}{6}\right)=cot\left(\dfrac{\Omega}{4}\right)\)
=>\(2x+\dfrac{\Omega}{6}=\dfrac{\Omega}{4}+k\Omega\)
=>\(2x=\dfrac{1}{12}\Omega+k\Omega\)
=>\(x=\dfrac{1}{24}\Omega+\dfrac{k\Omega}{2}\)
1) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}\)
2) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\dfrac{4n-\sqrt{16n^2+1}}{n+1}\right)\)
3) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}\right)\)
\(1,\lim\limits_{n\rightarrow\infty}\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}\left(1\right)\)
\(\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}=\dfrac{-\dfrac{n^2}{n^4}+\dfrac{2n}{n^4}+\dfrac{1}{n^4}}{\sqrt{\dfrac{3n^4}{n^4}+\dfrac{2}{n^4}}}=\dfrac{-\dfrac{1}{n^2}+\dfrac{2}{n^3}+\dfrac{1}{n^4}}{\sqrt{3+\dfrac{2}{n^4}}}\)
\(\Rightarrow\left(1\right)=\dfrac{-lim\dfrac{1}{n^2}+2lim\dfrac{1}{n^3}+lim\dfrac{1}{n^4}}{\sqrt{lim\left(3+\dfrac{2}{n^4}\right)}}\)
\(=\dfrac{0}{\sqrt{lim\left(3+\dfrac{2}{n^4}\right)}}=0\)
\(2,\lim\limits_{n\rightarrow\infty}\left(\dfrac{4n-\sqrt{16n^2+1}}{n+1}\right)\left(2\right)\)
\(\dfrac{4n-\sqrt{16n^2+1}}{n+1}=\dfrac{\dfrac{4n}{n^2}-\sqrt{\dfrac{16n^2}{n^2}+\dfrac{1}{n^2}}}{\dfrac{n}{n^2}+\dfrac{1}{n^2}}=\dfrac{\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}}{\dfrac{1}{n}+\dfrac{1}{n^2}}\)
\(\Rightarrow\left(2\right)=\dfrac{lim\left(\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}\right)}{lim\left(\dfrac{1}{n}+\dfrac{1}{n^2}\right)}=\dfrac{lim\left(\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}\right)}{0}\)
Vậy giới hạn \(\left(2\right)\) không xác định.
\(3,\lim\limits_{n\rightarrow\infty}\left(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}\right)\left(3\right)\)
\(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}=\dfrac{\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}}{\dfrac{2}{n}}\)
\(\Rightarrow\left(3\right)=\dfrac{lim\left(\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}\right)}{2lim\dfrac{1}{n}}=\dfrac{lim\left(\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}\right)}{0}\)
Vậy \(lim\left(3\right)\) không xác định.
Tìm các giới hạn sau:
\(\lim\limits_{x\rightarrow-\infty}\) \(\dfrac{\sqrt{x^6+2}}{3\text{x}^3-1}\)
\(\lim\limits_{x\rightarrow+\infty}\) \(\dfrac{\sqrt{x^6+2}}{3\text{x}^3-1}\)
\(\lim\limits_{x\rightarrow-\infty}\) \(\left(\sqrt{2\text{x}^2+1}+x\right)\)
\(\lim\limits_{x\rightarrow1}\) \(\dfrac{2\text{x}^3-5\text{x}-4}{\left(x+1\right)^2}\)
Tìm tập xác định của các hàm số sau:
1,\(y=sin\dfrac{3x+2}{2x-1}\)
2,\(y=tan\left(3x+\dfrac{2\pi}{5}\right)\)
3,\(y=cot\left(2x-\dfrac{1}{3}\right)\)
4,\(y=\dfrac{sinx+cosx}{sinx-cosx}\)
5,\(y=\dfrac{1}{sinx}+\dfrac{1}{cosx}\)
6,\(y=\dfrac{\sqrt{1-sinx}}{cosx}\)
7,\(y=\dfrac{3}{sin^2x-cos^2x}\)
8,\(y=\dfrac{1+tanx}{1+sinx}\)
9,\(y=\sqrt{\dfrac{1+sinx}{1-cosx}}\)
\(cosx-2cos3x=1+\sqrt{3}sinx\)
\(sinx+sinx\left(x+\dfrac{\pi}{3}\right)+sin4x=sin\left(2x-\dfrac{\pi}{3}\right)\)
\(\left(1-\dfrac{1}{2sinx}\right)cos^22x=2sinx-3+\dfrac{1}{sinx}\)
( sinx -2cosx)cos2x + sinx = (cos4x - 1)cosx +\(\dfrac{cos2x}{2sinx}\)
\(\left(\dfrac{cos4x+sin2x}{cos3x+sin3x}\right)^2=2\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+3\)
Giải các pt sau
a, \(\dfrac{1}{sinx}+\dfrac{1}{cosx}=4sin\left(x+\dfrac{\pi}{4}\right)\)
b, \(2sin\left(2x-\dfrac{\pi}{6}\right)+4sinx+1=0\)
c, \(cos2x+\sqrt{3}sinx+\sqrt{3}sin2x-cosx=2\)
d, \(4sin^2\dfrac{x}{2}-\sqrt{3}cos2x=1+cos^2\left(x-\dfrac{3\pi}{4}\right)\)
Tính các giới hạn sau:
a) \(\lim\limits_{x\rightarrow0^-}\dfrac{2\left|x\right|+x}{x^2-x}\)
b) \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\)
c) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt[3]{1+x^4+x^6}}{\sqrt{1+x^3+x^4}}\)
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: \(=lim_{x->-\infty}\left(\dfrac{x^2-x-x^2+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)
\(=lim_{x->-\infty}\left(\dfrac{-x+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)
\(=lim_{x->-\infty}\left(\dfrac{-1+\dfrac{1}{x}}{-\sqrt{1-\dfrac{1}{x^2}}-\sqrt{1-\dfrac{1}{x^2}}}\right)=\dfrac{-1}{-2}=\dfrac{1}{2}\)