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