1. **Stating the problem:** We need to evaluate the limit $$\lim_{x \to +\infty} (-3x^2 + 2\ln x - 2026)$$ and the definite integral $$\int_1^1 \left(\frac{1}{x} - 1\right) dx$$. 2. **Evaluating the limit:** The expression is $$-3x^2 + 2\ln x - 2026$$. As $$x \to +\infty$$, the term $$-3x^2$$ dominates because it grows faster than $$\ln x$$. Since $$-3x^2$$ tends to $$-\infty$$, the whole expression tends to $$-\infty$$. Therefore, $$ \lim_{x \to +\infty} (-3x^2 + 2\ln x - 2026) = -\infty $$ 3. **Evaluating the integral:** The integral is from 1 to 1 of $$\frac{1}{x} - 1$$. Recall that the definite integral from $$a$$ to $$a$$ of any function is zero because the interval length is zero. Therefore, $$ \int_1^1 \left(\frac{1}{x} - 1\right) dx = 0 $$ **Final answers:** - Limit: $$-\infty$$ - Integral: $$0$$