Question

Tính giới hạn

\(\overset{Lim}{x\rightarrow\dfrac{\pi}{6}}\dfrac{2\sin x-1}{6x-\pi}\)

Answer 1

Đặt \(x-\dfrac{\pi}{6}=t\Rightarrow x=t+\dfrac{\pi}{6}\)

\(\lim\limits_{t\rightarrow0}\dfrac{2sin\left(t+\dfrac{\pi}{6}\right)-1}{t}=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{3}sint+cost-1}{t}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{\sqrt{3}sint-2sin^2\dfrac{t}{2}}{t}=\lim\limits_{x\rightarrow0}\left(\sqrt{3}.\dfrac{sint}{t}-\dfrac{sin\dfrac{t}{2}}{\dfrac{t}{2}}.sin\dfrac{t}{2}\right)=\sqrt{3}.1-1.0=\sqrt{3}\)