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Câu a.$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{1-sinx}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{(cos\frac{x}{2}-sin\frac{x}{2})^2}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{2sin^2[\frac{1}{2}(\frac{\pi}{2}-x)]}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{\frac{1}{2}(\frac{\pi}{2}-x)^2.(\frac{sin[\frac{1}{2}(\frac{\pi}{2}-x]}{\frac{1}{2}(\frac{\pi}{2}-x)})^2}{(\frac{\pi}{2}-x)^2}$=$\frac{1}{2}$

Câu a.$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{1-sinx}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{(cos\frac{x}{2}-sin\frac{x}{2})^2}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{2sin^2[\frac{1}{2}(\frac{\pi}{2}-x)]}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{\frac{1}{2}(\frac{\pi}{2}-x)^2.[\frac{sin[\frac{1}{2}(\frac{\pi}{2}-x)}{\frac{1}{2}(\frac{\pi}{2}-x)}]^2}{(\frac{\pi}{2}-x)^2}$=$\frac{1}{2}$

Câu a.$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{1-sinx}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{(sin\frac{x}{2}-cos\frac{x}{2})^2}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{2sin^2[\frac{1}{2}(\frac{\pi}{2}-x)]}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{\frac{1}{2}(\frac{\pi}{2}-x)^2.(\frac{sin[\frac{1}{2}(\frac{\pi}{2}-x]}{\frac{1}{2}(\frac{\pi}{2}-x)})^2}{(\frac{\pi}{2}-x)^2}$=$\frac{1}{2}$

Câu a.$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{1-sinx}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{(cos\frac{x}{2}-sin\frac{x}{2})^2}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{2sin^2[\frac{1}{2}(\frac{\pi}{2}-x)]}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{\frac{1}{2}(\frac{\pi}{2}-x)^2.(\frac{sin[\frac{1}{2}(\frac{\pi}{2}-x]}{\frac{1}{2}(\frac{\pi}{2}-x)})^2}{(\frac{\pi}{2}-x)^2}$=$\frac{1}{2}$