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Câu hỏi: $\underset{x\to 0}{\mathop{\lim }} \dfrac{\cos \left( 3x \right)-1}{{{x}^{2}}}$ bằng
A. $\dfrac{9}{2}.$
B. $-\dfrac{3}{2}.$
C. $-\dfrac{2}{3}.$
D. $-\dfrac{9}{2}.$
A. $\dfrac{9}{2}.$
B. $-\dfrac{3}{2}.$
C. $-\dfrac{2}{3}.$
D. $-\dfrac{9}{2}.$
Cách 1: (Sử dụng giới hạn cơ bản)
$\underset{x\to 0}{\mathop{\lim }} \dfrac{\cos \left( 3x \right)-1}{{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }} \dfrac{-2{{\sin }^{2}}\dfrac{3x}{2}}{{{x}^{2}}}=-\dfrac{9}{2}\underset{x\to 0}{\mathop{\lim }} {{\left( \dfrac{\sin \dfrac{3x}{2}}{\dfrac{3x}{2}} \right)}^{2}}=-\dfrac{9}{2}$ (do $\underset{x\to 0}{\mathop{\lim }} \dfrac{\sin x}{x}=1$ ).
Cách 2: (Sử dụng quy tắc Lopital)
$\underset{x\to 0}{\mathop{\lim }} \dfrac{\cos \left( 3x \right)-1}{{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }} \dfrac{-3\sin \left( 3x \right)}{2x}=\underset{x\to 0}{\mathop{\lim }} \dfrac{-9\cos \left( 3x \right)}{2}=-\dfrac{9}{2}.$
$\underset{x\to 0}{\mathop{\lim }} \dfrac{\cos \left( 3x \right)-1}{{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }} \dfrac{-2{{\sin }^{2}}\dfrac{3x}{2}}{{{x}^{2}}}=-\dfrac{9}{2}\underset{x\to 0}{\mathop{\lim }} {{\left( \dfrac{\sin \dfrac{3x}{2}}{\dfrac{3x}{2}} \right)}^{2}}=-\dfrac{9}{2}$ (do $\underset{x\to 0}{\mathop{\lim }} \dfrac{\sin x}{x}=1$ ).
Cách 2: (Sử dụng quy tắc Lopital)
$\underset{x\to 0}{\mathop{\lim }} \dfrac{\cos \left( 3x \right)-1}{{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }} \dfrac{-3\sin \left( 3x \right)}{2x}=\underset{x\to 0}{\mathop{\lim }} \dfrac{-9\cos \left( 3x \right)}{2}=-\dfrac{9}{2}.$
Đáp án D.