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Đặt t=x -\pi thì khi x \to\pi ta có t \to 0
L=\mathop {\lim }\limits_{x \to\pi }\dfrac{\sin 3x}{\sin 5x}=\mathop {\lim }\limits_{t \to 0 }\dfrac{\sin 3(t+\pi)}{\sin 5(t+\pi)}=\mathop {\lim }\limits_{t \to 0 }\dfrac{\sin (3t+3\pi)}{\sin (5t+5\pi)}=\mathop {\lim }\limits_{t \to 0 }\dfrac{-\sin 3t}{-\sin 5t}=\mathop {\lim }\limits_{t \to 0 }\dfrac{\sin 3t}{\sin 5t}
L=\mathop {\lim }\limits_{t \to 0 }\left ( \dfrac{\sin 3t}{3t}.\dfrac{5t}{\sin 5t}.\dfrac{3}{5} \right )=1.1.\dfrac{3}{5}=\dfrac{3}{5}
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