mathop {lim }limits_{x to frac{pi }{2}} (1 - | vedclass

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(C) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o \frac{\pi }{2}} (1 - \sin x)	an x$.Rewrite the expression as: $\mathop {\lim }\limi

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(C) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o \frac{\pi }{2}} (1 - \sin x)	an x$.Rewrite the expression as: $\mathop {\lim }\limits_{x 	o \frac{\pi }{2}} (1 - \sin x) \frac{\sin x}{\cos x} = \mathop {\lim }\limits_{x 	o \frac{\pi }{2}} \frac{\sin x - \sin^2 x}{\cos x}$.As $x 	o \frac{\pi }{2}$,the expression takes the form $\frac{1-1}{0} = \frac{0}{0}$,which is an indeterminate form.Applying $L'	ext{Hospital's rule}$ by differentiating the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o \frac{\pi }{2}} \frac{\frac{d}{dx}(\sin x - \sin^2 x)}{\frac{d}{dx}(\cos x)} = \mathop {\lim }\limits_{x 	o \frac{\pi }{2}} \frac{\cos x - 2\sin x \cos x}{-\sin x}$.Substituting $x = \frac{\pi }{2}$:$\frac{\cos(\frac{\pi }{2}) - 2\sin(\frac{\pi }{2})\cos(\frac{\pi }{2})}{-\sin(\frac{\pi }{2})} = \frac{0 - 2(1)(0)}{-1} = \frac{0}{-1} = 0$.

$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} (1 - \sin x)\tan x$ is

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