\(1 - \sin x - \cos x = (1 - \cos x) - \sin x\)

\(=2\sin^2\left(\frac{x}{2}\right)-2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)=2\sin\left(\frac{x}{2}\right)\left[\sin\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right]\)

\(1 + \sin 3x + \cos 3x = (1 + \cos 3x) + \sin 3x\)

\(=2\cos^2\left(\frac{3x}{2}\right)+2\sin\left(\frac{3x}{2}\right)\cos\left(\frac{3x}{2}\right)=2\cos\left(\frac{3x}{2}\right)\left[\cos\left(\frac{3x}{2}\right)+\sin\left(\frac{3x}{2}\right)\right]\)

\(L = \lim_{x \to 0} \left( \frac{1 - \sin x - \cos x}{1 + \sin 3x + \cos 3x} \right)\)

\(=\lim_{x\to0}\frac{2\sin\left(\frac{x}{2}\right) \left[ \sin\left(\frac{x}{2}\right) - \cos\left(\frac{x}{2}\right) \right]}{2\cos\left(\frac{3x}{2}\right) \left[ \cos\left(\frac{3x}{2}\right) + \sin\left(\frac{3x}{2}\right) \right]}\)

\(=\lim_{x\to0}\left(\frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{3x}{2}\right)}\cdot\frac{\sin\left(\frac{x}{2}\right) - \cos\left(\frac{x}{2}\right)}{\cos\left(\frac{3x}{2}\right) + \sin\left(\frac{3x}{2}\right)}\right)\)

\(=\frac{0}{1}\cdot\frac{-1}{1}=0\cdot(-1)=0\)

Tui nghĩ cái này L'Hospital chứ giải thông thường là ko ổn :)

\(M=\lim\limits_{x\rightarrow0}\dfrac{\left(1+4x\right)^{\dfrac{1}{2}}-\left(1+6x\right)^{\dfrac{1}{3}}}{1-\cos3x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{2}\left(1+4x\right)^{-\dfrac{1}{2}}.4-\dfrac{1}{3}\left(1+6x\right)^{-\dfrac{2}{3}}.6}{3.\sin3x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-\dfrac{1}{4}.4\left(1+4x\right)^{-\dfrac{3}{2}}.4+\dfrac{2}{9}.6.6\left(1+6x\right)^{-\dfrac{5}{3}}}{3.3.\cos3x}\) 

Giờ thay x vô là được

\(N=\lim\limits_{x\rightarrow0}\dfrac{\left(1+ax\right)^{\dfrac{1}{m}}-\left(1+bx\right)^{\dfrac{1}{n}}}{\left(1+x\right)^{\dfrac{1}{2}}-1}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{m}.\left(1+ax\right)^{\dfrac{1}{m}-1}.a-\dfrac{1}{n}\left(1+bx\right)^{\dfrac{1}{n}-1}.b}{\dfrac{1}{2}\left(1+x\right)^{-\dfrac{1}{2}}}=\dfrac{\dfrac{a}{m}-\dfrac{b}{n}}{\dfrac{1}{2}}\)

\(V=\lim\limits_{x\rightarrow0}\dfrac{\left(1+mx\right)^n-\left(1+nx\right)^m}{\left(1+2x\right)^{\dfrac{1}{2}}-\left(1+3x\right)^{\dfrac{1}{3}}}=\lim\limits_{x\rightarrow0}\dfrac{n\left(1+mx\right)^{n-1}.m-m\left(1+nx\right)^{m-1}.n}{\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{1}{2}}.2-\dfrac{1}{3}\left(1+3x\right)^{-\dfrac{2}{3}}.3}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{n\left(n-1\right)\left(1+mx\right)^{n-2}.m-m\left(m-1\right)\left(1+nx\right)^{m-2}.n}{-\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{3}{2}}.2+\dfrac{2}{9}.3.3\left(1+3x\right)^{-\dfrac{5}{3}}}=....\left(thay-x-vo-la-duoc\right)\)

ĐKXĐ: ...

\(sin3x-cos3x+sinx+cosx=\dfrac{sin3x-cos3x+sinx+cosx}{\left(sin3x+cosx\right)\left(cos3x-sinx\right)}\)

\(\Rightarrow\left[{}\begin{matrix}sin3x-cos3x+sinx+cosx=0\left(1\right)\\\left(sin3x+cosx\right)\left(cos3x-sinx\right)=1\left(2\right)\end{matrix}\right.\)

(1) \(\Leftrightarrow3sinx-4sin^3x-4cos^3x+3cosx+sinx+cosx=0\)

\(\Leftrightarrow sinx+cosx+sin^3x+cos^3x=0\)

\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)\left(1-sinx.cosx\right)=0\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(2-sinx.cosx\right)=0\)

\(\Leftrightarrow sinx+cosx=0\) (loại)

(2) \(\Leftrightarrow sin3x.cos3x-sinx.cosx-sin3x.sinx+cos3x.cosx=1\)

\(\Leftrightarrow\dfrac{1}{2}sin6x-\dfrac{1}{2}sin2x+cos4x=1\)

\(\Leftrightarrow\dfrac{1}{2}\left(3sin2x-4sin^32x\right)-\dfrac{1}{2}sin2x+1-2sin^22x=1\)

\(\Leftrightarrow sin2x-2sin^32x-2sin^22x=0\)

\(\Leftrightarrow-sin2x\left(2sin^22x+2sin2x-1\right)=0\)

\(\Leftrightarrow...\)

\(\frac{1}{\sin x} - \frac{3}{\sin 3x} = \frac{1}{\sin x} - \frac{3}{\sin x (3 - 4\sin^2 x)}\)

\(= \frac{1}{\sin x} \left(1 - \frac{3}{3 - 4\sin^2 x}\right) = \frac{1}{\sin x} \left(\frac{3 - 4\sin^2 x - 3}{3 - 4\sin^2 x}\right)\)

\(= \frac{-4\sin^2 x}{\sin x (3 - 4\sin^2 x)} = \frac{-4\sin x}{3 - 4\sin^2 x}\)

\(L = \lim_{x \to 0} \left(\frac{-4\sin x}{3 - 4\sin^2 x}\right) \cdot \frac{1}{x} = \lim_{x \to 0} \left(\frac{\sin x}{x}\right) \cdot \frac{-4}{3 - 4\sin^2 x}\)

\(=1\cdot\frac{-4}{3 - 4(0)^2}=-\frac{4}{3}\)

\(1-\cos x\cdot\cos3x\)

\(=(1-\cos x)+(\cos x-\cos x\cdot\cos3x)\)

\(=(1-\cos x)+\cos x(1-\cos3x)\)

\(L = \lim_{x \to 0} \frac{1 - \cos x \cdot \cos 3x}{x^2}\)

\(=\lim_{x\to0}\frac{1 - \cos x}{x^2}+\lim_{x\to0}\left(\cos x\cdot\frac{1 - \cos3x}{x^2}\right)=L_1+L_2\)

\(L_1 = \lim_{x \to 0} \frac{2\sin^2\left(\frac{x}{2}\right)}{x^2} = \lim_{x \to 0} \frac{1}{2} \cdot \left[ \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}} \right]^2 = \frac{1}{2} \cdot 1^2 = \frac{1}{2}\)

\(\lim_{x \to 0} \frac{1 - \cos 3x}{x^2} = \lim_{x \to 0} \frac{2\sin^2\left(\frac{3x}{2}\right)}{x^2} = \lim_{x \to 0} 2 \cdot \frac{9}{4} \cdot \left[ \frac{\sin\left(\frac{3x}{2}\right)}{\frac{3x}{2}} \right]^2 = \frac{9}{2} \cdot 1^2 = \frac{9}{2}\)

Do đó: L=1/2+9/2=5

b: \(\Leftrightarrow2\cdot\cos2x\cdot\cos x+2\cdot\sin x\cdot\cos2x=\sqrt{2}\cdot\cos2x\)

\(\Leftrightarrow2\cdot\cos2x\left(\sin x+\cos x\right)=\sqrt{2}\cdot\cos2x\)

\(\Leftrightarrow\sqrt{2}\cdot\cos2x\cdot\left[\sqrt{2}\cdot\sqrt{2}\cdot\sin\left(x+\dfrac{\Pi}{4}\right)-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\cos2x=0\\\sin\left(x+\dfrac{\Pi}{4}\right)=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\Pi}{2}+k\Pi\\x+\dfrac{\Pi}{4}=\dfrac{\Pi}{6}+k2\Pi\\x+\dfrac{\Pi}{4}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\)

\(\Leftrightarrow x\in\left\{\dfrac{\Pi}{4}+\dfrac{k\Pi}{2};\dfrac{-1}{12}\Pi+k2\Pi;\dfrac{7}{12}\Pi+k2\Pi\right\}\)

c: \(\Leftrightarrow2\cdot\sin2x\cdot\cos x+\sin2x=2\cdot\cos2x\cdot\cos x+\cos2x\)

\(\Leftrightarrow\sin2x\left(2\cos x+1\right)=\cos2x\left(2\cos x+1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin2x=\cos2x=\sin\left(\dfrac{\Pi}{2}-2x\right)\\\cos x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{4}\\\\x=-\dfrac{2}{3}\Pi+k2\Pi\\x=\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)

\(a=\lim\limits_{x\rightarrow3}\dfrac{2x+3-x^2}{\left(x^2-4x+3\right)\left(\sqrt[]{2x+3}+x\right)}=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)\left(-x-1\right)}{\left(x-3\right)\left(x-1\right)\left(\sqrt[]{2x+3}+x\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{-x-1}{\left(x-1\right)\left(\sqrt[]{2x+3}+x\right)}=...\)

\(b=\lim\limits_{x\rightarrow0}\dfrac{\left(x+1\right)^{\dfrac{1}{3}}-1}{\left(2x+1\right)^{\dfrac{1}{4}}-1}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{3}\left(x+1\right)^{-\dfrac{2}{3}}}{\dfrac{1}{2}\left(2x+1\right)^{-\dfrac{3}{4}}}=\dfrac{2}{3}\)

\(c=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt[]{1+4x}-2x-1\right)+\left(2x+1-\sqrt[3]{1+6x}\right)}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{-4x^2}{2x+1+\sqrt[]{4x+1}}+\dfrac{x^2\left(8x+12\right)}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{-4}{2x+1+\sqrt[]{4x+1}}+\dfrac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}\right)=...\)

\(\frac{\left(sin3x+cosx\right)sin3x+\left(cos3x+sinx\right)cos3x}{cos4x}\)

\(=\frac{sin^23x+sin3x.cosx+cos^23x+cos3x.sinx}{cos4x}=\frac{1+sin3x.cosx+cos3x.sinx}{cos4x}\)

\(=\frac{1+sin4x}{cos4x}=\frac{sin^22x+cos^22x+2sin2x.cos2x}{cos^22x-sin^22x}=\frac{\left(cos2x+sin2x\right)^2}{\left(cos2x-sin2x\right)\left(cos2x+sin2x\right)}\)

\(=\frac{cos2x+sin2x}{cos2x-sin2x}=\frac{1+\frac{sin2x}{cos2x}}{1-\frac{sin2x}{cos2x}}=\frac{1+tan2x}{1-tan2x}\)

\(\lim\limits_{x\rightarrow0}\dfrac{\sin ax}{ax}=1\Rightarrow\sin ax\sim ax\Leftrightarrow\sin^2ax\sim\left(ax\right)^2\)

\(1-\cos x=1-\cos2.\dfrac{x}{2}=2\sin^2\dfrac{x}{2}\sim2.\left(\dfrac{x}{2}\right)^2=\dfrac{x^2}{2}\)

\(\Rightarrow\lim\limits_{x\rightarrow0}\dfrac{1-\cos2017x}{x^2}\)

Ta co khi \(x\rightarrow0:1-\cos2017x\sim\dfrac{\left(2017x\right)^2}{2}=\dfrac{2017^2x^2}{2}\)

\(\Rightarrow\lim\limits_{x\rightarrow0}\dfrac{1-\cos2017x}{x^2}=\lim\limits_{x\rightarrow0}\dfrac{2017^2x^2}{2x^2}=\dfrac{2017^2}{2}\)