1. **State the problem:** Find the limit $$\lim_{x \to 1} \frac{x + x^2 + x^3 + \cdots + x^{2025} - 2025}{x^2 - x}$$. 2. **Analyze the numerator:** The numerator is the sum of powers of $x$ from $x^1$ to $x^{2025}$ minus 2025: $$S = x + x^2 + \cdots + x^{2025} - 2025$$ Since the sum of a geometric series is: $$\sum_{k=1}^n x^k = x \frac{x^n - 1}{x - 1}$$ Applying this here, we get: $$\sum_{k=1}^{2025} x^k = x \frac{x^{2025} - 1}{x - 1}$$ So the numerator is: $$x \frac{x^{2025} - 1}{x - 1} - 2025$$ 3. **Rewrite the limit expression:** $$\lim_{x \to 1} \frac{x \frac{x^{2025} - 1}{x - 1} - 2025}{x^2 - x} = \lim_{x \to 1} \frac{x(x^{2025} - 1)/(x - 1) - 2025}{x^2 - x}$$ Multiply numerator and denominator by $(x-1)$ to clear the fraction inside numerator: $$= \lim_{x \to 1} \frac{x(x^{2025} - 1) - 2025(x-1)}{(x^2 - x)(x - 1)}$$ Note $x^2 - x = x(x-1)$, so denominator is: $$x(x-1)(x - 1) = x (x-1)^2$$ 4. **Rearrange the limit:** $$\lim_{x \to 1} \frac{x(x^{2025} - 1) - 2025(x-1)}{x (x-1)^2}$$ 5. **Define a function inside the numerator to apply L'Hôpital's rule or simplify:** Set $$f(x) = x(x^{2025} - 1) - 2025(x-1) = x^{2026} - x - 2025x + 2025 = x^{2026} - 2026 x + 2025$$ So the limit is: $$\lim_{x \to 1} \frac{f(x)}{x (x-1)^2}$$ 6. **Evaluate numerator and denominator at $x=1$:** $$f(1) = 1^{2026} - 2026\times1 + 2025 = 1 - 2026 + 2025 = 0$$ Denominator at $x=1$: $$1 \times (1-1)^2 = 0$$ Limit is indeterminate of form $0/0$. 7. **Use L'Hôpital's Rule:** Differentiate numerator and denominator with respect to $x$: Numerator derivative: $$f'(x) = 2026 x^{2025} - 2026$$ Denominator derivative: $$g(x) = x (x-1)^2$$ Use product rule: $$g'(x) = (x-1)^2 + x \times 2(x-1) = (x-1)^2 + 2x(x-1)$$ Simplify: $$(x-1)^2 + 2x(x-1) = (x-1)(x-1 + 2x) = (x-1)(3x - 1)$$ 8. **Evaluate derivatives at $x=1$:** $$f'(1) = 2026 \times 1^{2025} - 2026 = 2026 - 2026 = 0$$ $$g'(1) = (1-1)(3 \times 1 - 1) = 0$$ Still indeterminate $0/0$. 9. **Apply L'Hôpital's Rule second time:** Differentiate again. $$f''(x) = 2026 \times 2025 \times x^{2024} = 2026 \times 2025 x^{2024}$$ $$g'(x) = (x-1)(3x-1)$$ Differentiate $g'(x)$: $$g''(x) = (x-1) \times 3 + (3x - 1) \times 1 = 3(x-1) + (3x - 1) = 3x - 3 + 3x - 1 = 6x - 4$$ 10. **Evaluate second derivatives at $x=1$:** $$f''(1) = 2026 \times 2025 \times 1^{2024} = 2026 \times 2025$$ $$g''(1) = 6 \times 1 - 4 = 2$$ 11. **Final limit:** $$\lim_{x \to 1} \frac{f(x)}{g(x)} = \lim_{x \to 1} \frac{f''(x)}{g''(x)} = \frac{2026 \times 2025}{2}$$ Calculate the product: $$2026 \times 2025 = 2026 \times (2000 + 25) = 2026 \times 2000 + 2026 \times 25 = 4,052,000 + 50,650 = 4,102,650$$ Divide by 2: $$\frac{4,102,650}{2} = 2,051,325$$ **Answer:** $$\boxed{2,051,325}$$