1. **State the problem:** We need to find the limit $ \lim_{x \to 2} \frac{xe^{x-1} - 2e}{x-2} $. 2. **Recognize the form:** Substitute $x=2$ directly: $$\frac{2e^{2-1} - 2e}{2-2} = \frac{2e - 2e}{0} = \frac{0}{0}$$ This is an indeterminate form, so we can apply L'Hôpital's Rule. 3. **Apply L'Hôpital's Rule:** Differentiate numerator and denominator separately with respect to $x$: Numerator derivative: $$\frac{d}{dx} \left( xe^{x-1} - 2e \right) = \frac{d}{dx} \left( xe^{x-1} \right) - 0$$ Use product rule on $xe^{x-1}$: $$\frac{d}{dx} \left( x \right) \cdot e^{x-1} + x \cdot \frac{d}{dx} \left( e^{x-1} \right) = 1 \cdot e^{x-1} + x \cdot e^{x-1} = e^{x-1} + x e^{x-1} = e^{x-1}(1 + x)$$ Denominator derivative: $$\frac{d}{dx} (x-2) = 1$$ 4. **Rewrite the limit using derivatives:** $$\lim_{x \to 2} \frac{xe^{x-1} - 2e}{x-2} = \lim_{x \to 2} \frac{e^{x-1}(1 + x)}{1} = e^{2-1}(1 + 2) = e \cdot 3 = 3e$$ 5. **Final answer:** $$\boxed{3e}$$ This means the limit evaluates to $3e$.