1. **Stating the problem:** We need to find the following limits and evaluate the definite integral: A. $$\lim_{x \to +\infty} (3x - 5)$$ B. $$\lim_{x \to +\infty} e^x \left(-x^2 + \frac{3}{x}\right)$$ C. $$\lim_{x \to +\infty} x$$ (incomplete expression, so we skip this part) J. Evaluate the definite integral $$J = \int_0^3 \left(2x - \frac{1}{x+1}\right) dx$$ --- 2. **Solving A:** The limit of a linear function as $$x \to +\infty$$ is dominated by the term with the highest power of $$x$$. $$\lim_{x \to +\infty} (3x - 5) = \lim_{x \to +\infty} 3x - \lim_{x \to +\infty} 5 = +\infty - 5 = +\infty$$ So, the limit diverges to infinity. --- 3. **Solving B:** Consider the expression: $$\lim_{x \to +\infty} e^x \left(-x^2 + \frac{3}{x}\right)$$ Rewrite inside the parentheses: $$-x^2 + \frac{3}{x} = -x^2 + 3x^{-1}$$ As $$x \to +\infty$$, $$-x^2 \to -\infty$$ and $$3/x \to 0$$. Since $$e^x$$ grows faster than any polynomial, and the term inside parentheses tends to $$-\infty$$ dominated by $$-x^2$$, the product is: $$e^x \cdot (-x^2 + \frac{3}{x}) \approx e^x \cdot (-x^2) = -x^2 e^x$$ Since $$e^x$$ grows faster than $$x^2$$, the product tends to $$-\infty$$. Therefore: $$\lim_{x \to +\infty} e^x \left(-x^2 + \frac{3}{x}\right) = -\infty$$ --- 4. **Solving J:** Evaluate: $$J = \int_0^3 \left(2x - \frac{1}{x+1}\right) dx = \int_0^3 2x \, dx - \int_0^3 \frac{1}{x+1} \, dx$$ Calculate each integral separately: - $$\int_0^3 2x \, dx = \left[x^2\right]_0^3 = 3^2 - 0 = 9$$ - $$\int_0^3 \frac{1}{x+1} \, dx = \left[\ln|x+1|\right]_0^3 = \ln(4) - \ln(1) = \ln(4)$$ So, $$J = 9 - \ln(4)$$ --- **Final answers:** A. $$+\infty$$ B. $$-\infty$$ J. $$9 - \ln(4)$$