${\mathrm{let}}_{\mathrm{x}\to 1} \frac{\sqrt{1-\cos 2(\mathrm{x}-1)}}{\mathrm{x}-1}=$

$$\begin{gathered} Given limit=li{m}_{x\to 1}\frac{\sqrt{1-\cos 2(x-1)}}{x-1}=li{m}_{x\to 1}\frac{\sqrt{2{\sin }^2{(x-1)}^{ }}}{x-1} \\ = li{m}_{x\to 1}\frac{\sqrt{2}\left|\sin (x-1)\right|}{x-1} \\ Now {\mathrm{lt}}_{\mathrm{x}\to 1-} \mathrm{f}\left(\mathrm{x}\right)=li{m}_{x\to 1}\frac{\sqrt{2}(-\sin (x-1\left)\right)}{x-1}(\because x-1<0) \\ =-\sqrt{2}\left(1\right)=-\sqrt{2} \\ Now {{\mathrm{lt}}_{\mathrm{x}\to 1+}}^{ } \mathrm{f}\left(\mathrm{x}\right)=li{m}_{x\to 1}\frac{\sqrt{2}\left(\sin \right(x-1\left)\right)}{x-1}(\because x-1>0) \\ =\sqrt{2} \left(1\right) \\ =\sqrt{2} \\ \therefore {{\mathrm{lt}}_{\mathrm{x}\to 1+}}^{ } \mathrm{f}\left(\mathrm{x}\right)\ne {\mathrm{lt}}_{\mathrm{x}\to 1-} \mathrm{f}\left(\mathrm{x}\right) \\ \therefore limit does not exist \end{gathered}$$