\lim_{x \to 0^{-}}\frac{\ x+e^{ \frac{1}{x}}}{x} = 
\lim_{x \to 0^{-}}\frac{1-e^\frac{1}{x} \cdot \frac{1}{x^{2}}}{1} = 
1 - \lim_{x \to 0^{-}}e^\frac{1}{x} \cdot \frac{1}{x^{2}} = 
1 - \lim_{x \to 0^{-}}\frac{e^\frac{1}{x}}{x^{2}} = 
1 - \frac{e^{-\infty}}{-0^{2}} = 
1 - \frac{0}{0}