Answer
$$ - 1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1 + x}}{{{e^{ - x}} - 1 - x}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}0{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 0} \left( {{e^x} - 1 + x} \right) = {e^0} - 1 + 0 = 0 \cr
& \mathop {\lim }\limits_{x \to 0} \left( {{e^{ - x}} - 1 - x} \right) = {e^{ - 0}} - 1 - 0 = 0 \cr
& \cr
& {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. then differentiating the numerator and}} \cr
& {\text{denominator}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{D_x}\left( {{e^x} - 1 + x} \right)}}{{{D_x}\left( {{e^{ - x}} - 1 - x} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + 1}}{{ - {e^{ - x}} - 1}} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = \frac{{{e^0} + 1}}{{ - {e^{ - 0}} - 1}} = \frac{{1 + 1}}{{ - 1 - 1}} \cr
& = \frac{2}{{ - 2}} \cr
& = - 1 \cr
& {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1 + x}}{{{e^{ - x}} - 1 - x}} = - 1 \cr} $$
