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

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Question

(C) Given limit: $L = \mathop {\lim }\limits_{x 	o \pi /2} \frac{2x - \pi}{\cos x}$.Since the form is $\frac{0}{0}$,we apply $L$-Hospital's rule by d

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(C) Given limit: $L = \mathop {\lim }\limits_{x 	o \pi /2} \frac{2x - \pi}{\cos x}$.Since the form is $\frac{0}{0}$,we apply $L$-Hospital's rule by differentiating the numerator and denominator with respect to $x$:$L = \mathop {\lim }\limits_{x 	o \pi /2} \frac{\frac{d}{dx}(2x - \pi)}{\frac{d}{dx}(\cos x)}$$L = \mathop {\lim }\limits_{x 	o \pi /2} \frac{2}{-\sin x}$Substituting $x = \pi /2$:$L = \frac{2}{-\sin(\pi /2)} = \frac{2}{-1} = -2$.Thus,the correct option is $C$.

$\mathop {\lim }\limits_{x \to \pi /2} \frac{2x - \pi}{\cos x} = $

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