Tìm giới hạn \(\begin{equation} \lim\limits _{x \rightarrow 0} \frac{\tan x-\sin x}{\sin ^{3} x} \end{equation}\)
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Lời giải:
Báo saiTa có
\(\begin{array}{l} \mathop {\lim }\limits_{x \to 0} \frac{{\tan x - \sin x}}{{{{\sin }^3}x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x(1 - \cos x)}}{{\cos x{{\sin }^3}x}} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{\cos x{{\sin }^2}x}} = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}\frac{x}{2}}}{{\cos x{{\sin }^2}x}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{1}{{2\cos x}} \cdot \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{\sin \frac{x}{2}}}{{\frac{x}{2}}}} \right)^2} \cdot \mathop {\lim }\limits_{x \to 0} {\left( {\frac{x}{{\sin x}}} \right)^2} = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2} \end{array}\)
