f(x)= $\frac{a+b^{{\displaystyle 1/x}}}{c+d^{1/x}}, b>1, d>1, c\ne 0,$ f(0) = 1, is left continuous at x = 0, then

$$\begin{gathered} \mathrm{b}>1, \mathrm{d}>1, \mathrm{f}\left(0\right)=1. \\ \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{a}+{\mathrm{b}}^{1/\mathrm{x}}}{\mathrm{c}+{\mathrm{d}}^{1/\mathrm{x}}} \mathrm{is} \mathrm{left} \mathrm{continuous} \mathrm{at} \mathrm{x}=0 \\ \Rightarrow \underset{\mathrm{x}\to 0^{-}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=1\Rightarrow \underset{\mathrm{x}\to 0^{-}}{\lim }\frac{\mathrm{a}+{\mathrm{b}}^{1/\mathrm{x}}}{\mathrm{c}+{\mathrm{a}}^{1/\mathrm{x}}}=1 \\ \mathrm{Let} \mathrm{x}=0-\mathrm{h}, \mathrm{h}\to 0 \\ \Rightarrow \underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{a}+{\mathrm{b}}^{-{\displaystyle \frac{1}{\mathrm{h}}}}}{\mathrm{c}+{\mathrm{d}}^{-{\displaystyle \frac{1}{\mathrm{h}}}}}=1 \\ \therefore \frac{\mathrm{a}+0}{\mathrm{c}+0}=1\Rightarrow \mathrm{a}=\mathrm{c} \end{gathered}$$