The limit lim _{x rightarrow 1} frac{sqrt{1-c | vedclass

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

Question

(C) Let $L = \lim _{x ightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}$.Using the identity $1-\cos 2	heta = 2\sin^2	heta$,we have:$L = \lim _{x igh

Vedclass · Accepted Answer

does not exist

Vedclass · Answer

(C) Let $L = \lim _{x ightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}$.Using the identity $1-\cos 2	heta = 2\sin^2	heta$,we have:$L = \lim _{x ightarrow 1} \frac{\sqrt{2\sin^2(x-1)}}{x-1} = \sqrt{2} \lim _{x ightarrow 1} \frac{|\sin(x-1)|}{x-1}$.Let $z = x-1$. As $x ightarrow 1$,$z ightarrow 0$.$L = \sqrt{2} \lim _{z ightarrow 0} \frac{|\sin z|}{z}$.Now,evaluate the one-sided limits:Right-hand limit $(RHL)$: $\sqrt{2} \lim _{z ightarrow 0^+} \frac{\sin z}{z} = \sqrt{2}(1) = \sqrt{2}$.Left-hand limit $(LHL)$: $\sqrt{2} \lim _{z ightarrow 0^-} \frac{-\sin z}{z} = \sqrt{2}(-1) = -\sqrt{2}$.Since $RHL 
eq LHL$,the limit does not exist.

The limit $\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}$

Similar Questions

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} x \sec x$

$\lim _{x \rightarrow 0} \frac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2} = $

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{\log (1 + {x^3})}}{{{{\sin }^3}x}} = $

$\lim _{n \rightarrow \infty}\left[\frac{1^3}{1-n^4}+\frac{2^3}{1-n^4}+\ldots +\frac{n^3}{1-n^4}\right]=$

$\mathop {\lim }\limits_{x \to 2} \frac{|x - 2|}{x - 2} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module