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  • <img alt="\mathrm{\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightarrow%201%7D%20%5Cfrac%7B%5Csq

\mathrm{\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}}
 

Option: 1

exists and it equals \sqrt{2}


Option: 2

exists and it equals -\sqrt{2}
 


Option: 3

doesnot exist because \mathrm{(x-1) \rightarrow 0}
 


Option: 4

doesnot exist because left hand limit is not equal to right hand limit.


Answers (1)

best_answer

\mathrm{\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}=\lim _{x \rightarrow 1} \frac{\sqrt{2 \sin ^2(x-1)}}{x-1}}

                                            \mathrm{ =\lim _{x \rightarrow 1} \sqrt{2} \frac{|\sin (x-1)|}{x-1} }

\mathrm{ \therefore(\text { LHL at } x=1) =\lim _{x \rightarrow 1^{-}} \sqrt{2} \frac{|\sin (x-1)|}{x-1} }

                                   \mathrm{ =\lim _{x \rightarrow 1}-\sqrt{2} \frac{\sin (x-1)}{x-1}=-\sqrt{2} }

\mathrm{ and, (RHL\: at \: x=1)=\lim _{x \rightarrow 1^{+}} \sqrt{2} \frac{\sin (x-1) \mid}{x-1} }

                                      \mathrm{ =\lim _{x \rightarrow 1} \sqrt{2} \frac{\sin (x-1)}{x-1}=\sqrt{2} }

Clearly, \mathrm{ ( LHL\: at \: x=1) \neq( RHL\: at \: x=1) }

So, \mathrm{ \lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1} } does not exist.

Hence option 4 is correct.

Posted by

mansi

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