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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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} = $

Similar Questions

If $\alpha = \lim_{x \rightarrow 0} \frac{x \cdot 2^x - x}{1 - \cos x}$ and $\beta = \lim_{x \rightarrow 0} \frac{x \cdot 2^x - x}{\sqrt{1+x^2} - \sqrt{1-x^2}}$,then

$\mathop {\lim }\limits_{x \to 0} \frac{{\tan 2x - x}}{{3x - \sin x}} = $

$\operatorname{Lim}_{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$ is equal to :

$\lim _{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$ is equal to

If $f(9) = 9$ and $f'(9) = 4$,then $\mathop {\lim }\limits_{x \to 9} \frac{{\sqrt {f(x)} - 3}}{{\sqrt x - 3}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module