If 0 leq x leq pi / 2,then lim _{x rightarrow | vedclass

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

Question

(D) Let $f(x) = \frac{|2 \cos x - 1|}{2 \cos x - 1}$ for $0 \leq x \leq \pi/2$. We know that $2 \cos x - 1 > 0$ when $\cos x > 1/2$,i.e.,$0 \leq x < \

Vedclass · Accepted Answer

$=1$,when $0 \leq a < \pi/3$

Vedclass · Answer

(D) Let $f(x) = \frac{|2 \cos x - 1|}{2 \cos x - 1}$ for $0 \leq x \leq \pi/2$. We know that $2 \cos x - 1 > 0$ when $\cos x > 1/2$,i.e.,$0 \leq x Thus,$f(x) = 1$ for $0 \leq x For the limit to exist at $x=a$,the left-hand limit and right-hand limit must be equal. For $0 \leq a For $\pi/3 At $a = \pi/3$,the left-hand limit is $1$ and the right-hand limit is $-1$,so the limit does not exist. Therefore,the correct statement is that the limit is $1$ when $0 \leq a < \pi/3$.

If $0 \leq x \leq \pi / 2$,then $\lim _{x \rightarrow a} \frac{|2 \cos x-1|}{2 \cos x-1}$

Similar Questions

Evaluate the given limit: $\mathop {\lim }\limits_{r \to 1} \pi r^{2}$

If $\alpha$ and $\beta$ are the roots of the equation $375x^2 - 25x - 2 = 0$,then $\lim_{n \to \infty} \sum_{r=1}^n \alpha^r + \lim_{n \to \infty} \sum_{r=1}^n \beta^r$ is equal to

$\mathop {Limit}\limits_{x \to \frac{\pi }{2}} \,\frac{{\sin x}}{{{{\cos }^{ - 1}}\left[ {\frac{1}{4}\,(3\sin x\, - \,\sin 3x)} \right]}}\,$,where $[ \cdot ]$ denotes the greatest integer function,is

$\lim _{x \rightarrow 0} \frac{(\operatorname{cosec} x-\cot x)(e^x-e^{-x})}{\sqrt{3}-\sqrt{2+\cos x}} = $

Evaluate the limit: $\lim_{x \rightarrow 0} \left( \frac{4^x - 1}{2^x - 1} - \frac{\sqrt{4 + 3x} - 2}{x} \right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module