## Calculus with Applications (10th Edition)

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\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}