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: $$\frac{2e^2 - 1 - 2e}{2 - 2} = \frac{2e^2 - 1 - 2e}{0}$$ which is an indeterminate form of type $\frac{0}{0}$ if $2e^2 - 1 - 2e = 0$. 3. **Check numerator at $x=2$:** Calculate $2e^2 - 1 - 2e$: $$2e^2 - 2e - 1$$ We need to verify if this equals zero. 4. **Apply L'Hôpital's Rule:** Since direct substitution gives $\frac{0}{0}$, differentiate numerator and denominator separately: - Numerator derivative: $$\frac{d}{dx}(xe^x - 1 - 2e) = e^x + xe^x - 0 - 0 = e^x(1 + x)$$ - Denominator derivative: $$\frac{d}{dx}(x - 2) = 1$$ 5. **Evaluate the limit of the derivatives at $x=2$:** $$\lim_{x \to 2} \frac{e^x(1 + x)}{1} = e^2(1 + 2) = 3e^2$$ 6. **Final answer:** $$\boxed{3e^2}$$