# Problem Evaluate the limit: \[ \lim_{x \to } \frac{e^{2x} - 1 - 2x}{x^2} \] # Step-by-Step Solution First, observe that as \( x \to \), both the numerator and denominator approach zero, so the limit is an indeterminate form \( \frac{}{} \). This suggests we can apply L'Hôpital's Rule or expand the numerator using a Taylor series. Let's expand \( e^{2x} \) as a Taylor series around \( x = \): \[ e^{2x} = 1 + 2x + 2x^2 + \frac{4x^3}{3!} + \cdots \] Now substitute this expansion into the numerator: \[ e^{2x} - 1 - 2x = (1 + 2x + 2x^2 + \frac{4x^3}{3!} + \cdots) - 1 - 2x = 2x^2 + \frac{4x^3}{6} + \cdots \] So, \[ \frac{e^{2x} - 1 - 2x}{x^2} = \frac{2x^2 + \frac{2}{3}x^3 + \cdots}{x^2} = 2 + \frac{2}{3}x + \cdots \] As \( x \to \), the higher order terms vanish, and the limit approaches 2. # Final Answer \[ \boxed{2} \]