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 $\mathop {\lim }\limits_{x \to 0}\frac{x}{\sqrt{1-\cos x}}$
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$
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 $\mathop {\lim }\limits_{x \to 0}\frac{x}{\sqrt{1-\cos x}}$