1. **State the problem:** We are given the limit $$\lim_{x \to 1} \frac{h(x) - h(1)}{\ln(x^2)} = 2$$ and asked to find the value of $$h'(1)$$ and analyze if $$h(1)$$ is an extremum. 2. **Recall the definition of derivative:** The derivative of a function $$h$$ at $$x=1$$ is defined as $$ h'(1) = \lim_{x \to 1} \frac{h(x) - h(1)}{x - 1} $$ 3. **Rewrite the denominator:** Note that $$\ln(x^2) = 2 \ln x$$. As $$x \to 1$$, $$\ln x \to 0$$, so the denominator approaches 0. 4. **Relate the given limit to the derivative:** The limit given is similar to the derivative definition but with $$\ln(x^2)$$ instead of $$x-1$$. To connect them, use the substitution: $$ \lim_{x \to 1} \frac{h(x) - h(1)}{\ln(x^2)} = \lim_{x \to 1} \frac{h(x) - h(1)}{x - 1} \cdot \frac{x - 1}{\ln(x^2)} $$ 5. **Evaluate the second factor:** Use the limit $$ \lim_{x \to 1} \frac{x - 1}{\ln(x^2)} $$ Apply L'Hôpital's Rule since numerator and denominator both approach 0: $$ \lim_{x \to 1} \frac{1}{\frac{d}{dx} \ln(x^2)} = \lim_{x \to 1} \frac{1}{\frac{2}{x}} = \frac{1}{2} $$ 6. **Combine the limits:** Since the original limit equals 2, $$ 2 = h'(1) \times \frac{1}{2} \implies h'(1) = 4 $$ 7. **Check if $$h(1)$$ is an extremum:** For $$h(1)$$ to be an extremum, $$h'(1)$$ must be 0. Since $$h'(1) = 4 \neq 0$$, $$h(1)$$ is not an extremum. **Final answers:** - (A) $$h'(1) = 2$$ is false. - (B) $$h'(1) = 4$$ is true. - (C) $$h'(1) = 1$$ is false. - (D) $$h(1)$$ is an extremum is false. --- Regarding the function $$f(x) = (2x^2 + 6x + 5) e^{-2x}$$, it is given but not directly related to the limit problem above, so no further action is needed here.