As $x \to 2^-$ , $\dfrac{1}{x-2} \to -\infty$( so $e^{\frac{1}{x-2}} \to 0 $) , so $\lim_{x\to 2^-} \dfrac{1}{1+e^{\frac{1}{x-2}}} = \dfrac{1}{1+0} = 1$
As $x \to 2^+$ , $\dfrac{1}{x-2} \to \infty$ ( so $e^{\frac{1}{x-2} }\to \infty $) , so $\lim_{x\to 2^+} \dfrac{1}{1+e^{\frac{1}{x-2}}} = 0 $
Left Hand limit $\neq$ Right hand limit, so limit DNE , option D is correct one.