mathop {lim }limits_{x to 1} frac{{sqrt {1 - | vedclass

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

Question

(D) Let $f(x) = \frac{\sqrt{1 - \cos 2(x - 1)}}{x - 1}$.We know that $1 - \cos 2	heta = 2\sin^2 	heta$,so $1 - \cos 2(x - 1) = 2\sin^2(x - 1)$.Thus,

Vedclass · Accepted Answer

Does not exist because left hand limit is not equal to right hand limit

Vedclass · Answer

(D) Let $f(x) = \frac{\sqrt{1 - \cos 2(x - 1)}}{x - 1}$.We know that $1 - \cos 2	heta = 2\sin^2 	heta$,so $1 - \cos 2(x - 1) = 2\sin^2(x - 1)$.Thus,$f(x) = \frac{\sqrt{2\sin^2(x - 1)}}{x - 1} = \frac{\sqrt{2}|\sin(x - 1)|}{x - 1}$.Right Hand Limit $(RHL)$: $\mathop {\lim }\limits_{x 	o 1^+} f(x) = \mathop {\lim }\limits_{h 	o 0^+} \frac{\sqrt{2}|\sin h|}{h} = \mathop {\lim }\limits_{h 	o 0^+} \sqrt{2} \frac{\sin h}{h} = \sqrt{2}$.Left Hand Limit $(LHL)$: $\mathop {\lim }\limits_{x 	o 1^-} f(x) = \mathop {\lim }\limits_{h 	o 0^+} \frac{\sqrt{2}|\sin(-h)|}{-h} = \mathop {\lim }\limits_{h 	o 0^+} \frac{\sqrt{2}\sin h}{-h} = -\sqrt{2}$.Since $	ext{RHL} 
eq 	ext{LHL}$,the limit does not exist.

$\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {1 - \cos 2(x - 1)} }}{{x - 1}}$

Similar Questions

$\lim _{x \rightarrow 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin ^2 x} = $

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 2} \frac{3x^{2}-x-10}{x^{2}-4}$

$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{2}{x}} \right)^x} = $

$\lim_{x \rightarrow -\infty} \frac{3|x|-x}{|x|-2x} - \lim_{x \rightarrow 0} \frac{\log(1+x^3)}{\sin^3 x} =$

$\mathop {\lim }\limits_{x \to 0} \cos \frac{1}{x}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module