Question

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

• A. $a=1,b=e^{-1}$
• B. $a=2,b=e^{-1}$
• C. $a=-1,b=e^{-1}$
• D. $a=1,b=0$

Answer 1

The correct option is D $a=1,b=0$

We have,
$a\cdot \underset{x\to 1}{lim}{x}^{\frac{1}{1-x}}+b=e^{-1}$
$\Rightarrow a\cdot \underset{x\to 1}{lim}[1+(x-1){]}^{\frac{1}{1-x}}+b=e^{-1}$
$\Rightarrow a{e}^{\underset{x\to 1}{lim}\frac{x-1}{1-x}}+b=e^{-1}$
$\Rightarrow a\cdot e^{-1}+b=e^{-1}\Rightarrow a=1$ and $b=0$
Hence, option 'D' is correct.