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Lời giải:

Ta có:

Áp dụng công thức lượng giác: \(\sin (a-b)=\sin a\cos b-\cos a\sin b\)

thì:

\(\sqrt{3}\sin x-\cos x=-2\left(\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x\right)=-2\left(\sin \frac{\pi}{6}\cos x-\cos \frac{\pi}{6}\sin x\right)\)

\(=-2\sin \left(\frac{\pi}{6}-x\right)\)

Do đó: \(\lim_{x\to \frac{\pi}{6}}\frac{\sqrt{3}\sin x-\cos x}{\sin (\frac{\pi}{3}-2x)}=-2\lim_{x\to \frac{\pi}{6}}\frac{\sin \left ( \frac{\pi}{6}-x \right )}{\sin \left [ 2(\frac{\pi}{6}-x) \right ]}\)

\(=-\lim_{x\to \frac{\pi}{6}}\frac{\sin \left ( \frac{\pi}{6}-x \right )}{\frac{\pi}{6}-x}.\lim_{x\to \frac{\pi}{6}}\frac{1}{\frac{\sin\left [ 2(\frac{\pi}{6}-x) \right ]}{2(\frac{\pi}{6}-x)}}=-1.1.1=-1\)

(sử dụng công thức \(\lim_{t\to 0} \frac{\sin t}{t}=1\) . Trong TH bài toán \(x\to \frac{\pi}{6}\Rightarrow \frac{\pi}{6}-x\to 0\) )

\(\lim\limits_{x\rightarrow a}\frac{sin\left(\frac{x-a}{2}\right)}{\frac{x-a}{2}}.cos\left(\frac{x+a}{2}\right)=1.cos\left(\frac{a+a}{2}\right)=cosa\)

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:

\(\lim\limits_{x\rightarrow a}\frac{\left(tanx-tana\right)\left(tanx+tana\right)}{\frac{sin\left(x-a\right)}{cos\left(x-a\right)}}=\lim\limits_{x\rightarrow a}\frac{sin\left(x-a\right)\left(tanx+tana\right).cos\left(x-a\right)}{sin\left(x-a\right).cosx.cosa}=\lim\limits_{x\rightarrow a}\frac{\left(tanx+tana\right).cos\left(x-a\right)}{cosx.cosa}\)

\(=\frac{2tana}{cos^2a}\)

a/ \(\lim\limits_{x\to 1} f(x)=\frac{x^{2}-5x + 6}{x-2} \)

\(<=>\lim\limits_{x\to 1} f(x)=\dfrac{(x-3)(x-2)}{x-2} \)

<=>\(\lim\limits_{x\to 1} f(x)=x-3 \)

\(<=>\lim\limits_{x\to 1} f(x)=-2\)

a.

\(y'=\dfrac{3}{cos^2\left(3x-\dfrac{\pi}{4}\right)}-\dfrac{2}{sin^2\left(2x-\dfrac{\pi}{3}\right)}-sin\left(x+\dfrac{\pi}{6}\right)\)

b.

\(y'=\dfrac{\dfrac{\left(2x+1\right)cosx}{2\sqrt{sinx+2}}-2\sqrt{sinx+2}}{\left(2x+1\right)^2}=\dfrac{\left(2x+1\right)cosx-4\left(sinx+2\right)}{\left(2x+1\right)^2}\)

c.

\(y'=-3sin\left(3x+\dfrac{\pi}{3}\right)-2cos\left(2x+\dfrac{\pi}{6}\right)-\dfrac{1}{sin^2\left(x+\dfrac{\pi}{4}\right)}\)

Đặt \(t=x-\dfrac{\pi}{4}\), khi đó:

\(\lim\limits_{x\rightarrow\dfrac{\pi}{4}}\dfrac{\sqrt{2}cosx-1}{\sqrt{2}sinx-1}=\lim\limits_{t\rightarrow0}\dfrac{\sqrt{2}cos\left(t+\dfrac{\pi}{4}\right)-1}{\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)-1}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{cost-sint-1}{cost+sint-1}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{1-2sin^2\dfrac{t}{2}-2sin\dfrac{t}{2}.cos\dfrac{t}{2}-1}{1-2sin^2\dfrac{t}{2}+2sin\dfrac{t}{2}.cos\dfrac{t}{2}-1}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{-2sin\dfrac{t}{2}\left(sin\dfrac{t}{2}+cos\dfrac{t}{2}\right)}{-2sin\dfrac{t}{2}\left(sin\dfrac{t}{2}-cos\dfrac{t}{2}\right)}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{sin\dfrac{t}{2}+cos\dfrac{t}{2}}{sin\dfrac{t}{2}-cos\dfrac{t}{2}}\)

\(=-1\)

L'Hospital đi em

lim đề bài cho = \(\lim\limits_{x->\dfrac{\pi}{4}}\dfrac{-\sqrt{2}sinx}{\sqrt{2}cosx}\)

Thay x vào là xong