1. **State the problem:** Find the limit $$\lim_{x \to 0} x \frac{f(5+x) - f(5)}{x}$$ where $$f(x) = 2x^2 + 3$$. 2. **Rewrite the expression:** The limit can be simplified as $$\lim_{x \to 0} \frac{f(5+x) - f(5)}{\cancel{x}} \times \cancel{x} = \lim_{x \to 0} (f(5+x) - f(5))$$ because the $$x$$ in numerator and denominator cancel. 3. **Calculate $$f(5+x)$$:** $$f(5+x) = 2(5+x)^2 + 3 = 2(25 + 10x + x^2) + 3 = 50 + 20x + 2x^2 + 3 = 53 + 20x + 2x^2$$ 4. **Calculate $$f(5)$$:** $$f(5) = 2(5)^2 + 3 = 2(25) + 3 = 50 + 3 = 53$$ 5. **Substitute into the difference:** $$f(5+x) - f(5) = (53 + 20x + 2x^2) - 53 = 20x + 2x^2$$ 6. **Substitute back into the limit:** $$\lim_{x \to 0} x \frac{20x + 2x^2}{x} = \lim_{x \to 0} (20x + 2x^2)$$ 7. **Evaluate the limit:** $$\lim_{x \to 0} (20x + 2x^2) = 20 \times 0 + 2 \times 0^2 = 0$$ **Final answer:** $$0$$