A) exists and it equals \[\sqrt{2}\]
B) exists and it equals \[-\sqrt{2}\]
C) does not exist because \[x-1\to 0\]
D) does not exist because left hand limit is not equal to right hand limit
Correct Answer: D
Solution :
LHL\[=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\frac{\sqrt{1-\cos 2(x-1)}}{x-1}\] \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\frac{\sqrt{2-{{\sin }^{2}}(x-1)}}{x-1}\] \[=\sqrt{2}\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\frac{|\sin (x-1)|}{x-1}\] Put\[x=1-h,h>0,\] For\[x\to {{1}^{-}},h\to 0\] \[=\sqrt{2}\underset{x\to 0}{\mathop{\lim }}\,\frac{|\sin (-h)|}{-h}\] \[=\sqrt{2}\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin h}{-h}=-\sqrt{2}\] Again RHL\[=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\frac{\sqrt{1-\cos 2(x-1)}}{x-1}\] \[=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\sqrt{2}\frac{|\sin (x-1)|}{x-1}\] Put\[x=1,+h,h>0\] For\[x\to {{1}^{+}},h\to 0\] \[=\underset{h\to 0}{\mathop{\lim }}\,\sqrt{2}\frac{|\sinh |}{h}=\underset{x\to 0}{\mathop{\lim }}\,\sqrt{2}\frac{\sinh }{h}=\sqrt{2}\] \[LHL\ne RHL\] Therefore, \[\underset{x\to 1}{\mathop{\lim }}\,f(x)\] does not exist.
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