Question

lt x tends to 0 xe^x/1-e^x​

Answer 1

$\lim\limits_{x\to 0}\dfrac{xe^x}{1-e^x} = \lim\limits_{x\to 0}\dfrac{xe^x-x+x}{1-e^x} =\\ \\ =\lim\limits_{x\to 0}\dfrac{x(e^x-1)}{1-e^x}+\lim\limits_{x\to 0}\dfrac{x}{1-e^x} =\\ \\ = 0+\lim\limits_{x\to 0}\dfrac{x}{1-e^x} \overset{\frac{0}{0}(L'H.)}{=}\lim\limits_{x\to 0}\dfrac{1}{-e^x}=\\ \\ =\boxed{-1}$

Answer 2

Answer:

-1

Step-by-step explanation:

xe^x

-----------

1-e^x​

     e^x

= -----------

  - (e^x - 1​)/x

= 1/(-1)

= -1