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Giới hạn cơ bản: \mathop {\lim }\limits_{x \to 0}\frac{\sin x}{x}=\mathop {\lim }\limits_{x \to 0}\frac{x}{\sin x}=1
Ta có:
\mathop {\lim }\limits_{x \to 0^+}\frac{x}{\sqrt{1-\cos x}}
=\mathop {\lim }\limits_{x \to 0^+}\frac{x}{\sqrt2\sin\frac{x}{2}}
=\mathop {\lim }\limits_{x \to 0^+}\frac{\sqrt2.\frac{x}{2}}{\sin\frac{x}{2}}=\sqrt2
Mặt khác:
\mathop {\lim }\limits_{x \to 0^-}\frac{x}{\sqrt{1-\cos x}}
=\mathop {\lim }\limits_{x \to 0^-}\frac{x}{-\sqrt2\sin\frac{x}{2}}
=\mathop {\lim }\limits_{x \to 0^-}\frac{-\sqrt2.\frac{x}{2}}{\sin\frac{x}{2}}=-\sqrt2
Suy ra không tồn tại \mathop {\lim }\limits_{x \to 0}\frac{x}{\sqrt{1-\cos x}}
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