Question

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

$a\underset{x\to 1}{lim}(x^{\frac{1}{1-x}}+b)$

Put $x-1=h$

$x=1+h$

as $x\to 1,h\to 0$

$a\underset{h\to 0}{lim}{(1+h)}^{\frac{1}{-h}}+b=e^{-1}$

$a{\left\{\underset{h\to 0}{lim}{(1+h)}^{\frac{1}{h}}\right\}}^{-1}+b=e^{-1}$

Apply the common limit: $\underset{h\to 0}{lim}\left({(1+h)}^{\frac{1}{h}}\right)=e$

$a{e}^{-1}+b=e^{-1}$

$a=1,b=0$