If $\text{ a }\underset{x\to 1}{lim} x^{\frac{1}{1-x}}+b=e^{-1}(a\ge 1,b\ge 0)$ then

Solution:

We have,

$\begin{array}{l}a\cdot \underset{x\to 1}{lim} x^{1/1-x}+b=e^{-1}\\ \Rightarrow a\cdot \underset{x\to 1}{lim} [1+(x-1){]}^{1/1-x}+b=e^{-1}\\ \Rightarrow a\underset{x\to 1}{lim} e^{\frac{x-1}{1-x}}+b=e^{-1}\\ \Rightarrow a e^{-1}+b=e^{-1}\Rightarrow a=1\text{ and }b=0\end{array}$