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$L=\mathop {\lim }\limits_{x \to \frac{\pi }{2}}\frac{1-\sin x}{(\dfrac{\pi }{2}-x)^2}$
Đặt $t=\dfrac{\pi }{2}-x\Rightarrow $ khi $x \to \frac{\pi }{2}$ thì $t \to 0^-$. Ta có
$L=\mathop {\lim }\limits_{t \to 0^-}\frac{1-\sin (\dfrac{\pi }{2}-t)}{t^2}=\mathop {\lim }\limits_{t \to 0^-}\frac{1-\cos t}{t^2}=\mathop {\lim }\limits_{t \to 0^-} \dfrac{1}{2}\frac{\sin^2 \frac{t}{2}}{\frac{t^2}{4}}= \dfrac{1}{2}.$
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