Step-by-step Solution:

Apply L'Hôpital's Rule \[\] L'Hôpital's Rule states that for limits of the form \( \frac{0}{0} \), we can differentiate the numerator and denominator separately and then take the limit of the resulting fraction. \[\] Numerator: Differentiate \( e^x - e^{-x} - 2x \) \[ \frac{d}{dx}(e^x - e^{-x} - 2x) = e^x + e^{-x} - 2. \] Denominator: Differentiate \( 1 - \cos x \) \[ \frac{d}{dx}(1 - \cos x) = \sin x. \] Thus, the new expression after applying L'Hôpital's Rule is: \[ \lim_{x \to 0} \frac{e^x + e^{-x} - 2}{\sin x}. \] Apply L'Hôpital's Rule again \[\] Numerator: Differentiate \( e^x + e^{-x} - 2 \) \[ \frac{d}{dx}(e^x + e^{-x} - 2) = e^x - e^{-x}. \] Denominator: Differentiate \( \sin x \) \[ \frac{d}{dx}(\sin x) = \cos x. \] The new expression after applying L'Hôpital's Rule again is: \[ \lim_{x \to 0} \frac{e^x - e^{-x}}{\cos x}. \] Evaluate the limit as \( x \to 0 \) Substitute \( x = 0 \) into the new expression: Numerator: \( e^0 - e^0 = 1 - 1 = 0 \). Denominator: \( \cos(0) = 1 \). The limit is: \[ \frac{0}{1} = 0. \] Final Answer: \[ \lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{1 - \cos x} = 0. \]