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(D) Let $g(x) = ax + b$. Since $f(x)$ is continuous at $x = 0$,we have $\lim_{x 	o 0^+} f(x) = f(0)$.$\lim_{x 	o 0^+} \left(\frac{1+x}{2+x}ight)^{

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$\log_e\left(\frac{4}{9 e^{1/3}}\right)$

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(D) Let $g(x) = ax + b$. Since $f(x)$ is continuous at $x = 0$,we have $\lim_{x 	o 0^+} f(x) = f(0)$.$\lim_{x 	o 0^+} \left(\frac{1+x}{2+x}ight)^{\frac{1}{x}} = g(0) = b$.Evaluating the limit: $\lim_{x 	o 0^+} \left(\frac{1+x}{2+x}ight)^{\frac{1}{x}} = e^{\lim_{x 	o 0} \frac{1}{x} \ln\left(\frac{1+x}{2+x}ight)} = e^{\lim_{x 	o 0} \frac{1}{x} (\ln(1+x) - \ln(2+x))} = e^{\lim_{x 	o 0} \frac{1}{x} (x - \ln 2 - \frac{x}{2})} = e^{-\infty} = 0$.Thus,$b = 0$,so $g(x) = ax$.For $x > 0$,$f(x) = \left(\frac{1+x}{2+x}ight)^{\frac{1}{x}}$. Let $y = f(x)$,then $\ln y = \frac{1}{x} (\ln(1+x) - \ln(2+x))$.$\frac{1}{y} f'(x) = -\frac{1}{x^2} (\ln(1+x) - \ln(2+x)) + \frac{1}{x} \left(\frac{1}{1+x} - \frac{1}{2+x}ight)$.At $x = 1$,$y = f(1) = \frac{2}{3}$.$\frac{3}{2} f'(1) = -(\ln 2 - \ln 3) + 1 \left(\frac{1}{2} - \frac{1}{3}ight) = \ln\left(\frac{3}{2}ight) + \frac{1}{6}$.$f'(1) = \frac{2}{3} \ln\left(\frac{3}{2}ight) + \frac{1}{9} = -\frac{2}{3} \ln\left(\frac{2}{3}ight) + \frac{1}{9}$.Given $f'(1) = f(-1) = g(-1) = -a$,so $a = \frac{2}{3} \ln\left(\frac{2}{3}ight) - \frac{1}{9}$.$g(3) = 3a = 2 \ln\left(\frac{2}{3}ight) - \frac{1}{3} = \ln\left(\left(\frac{2}{3}ight)^2ight) - \ln(e^{1/3}) = \ln\left(\frac{4}{9 e^{1/3}}ight)$.

Let $g(x)$ be a linear function and $f(x) = \begin{cases} g(x) & , x \leq 0 \\ \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}} & , x > 0 \end{cases}$ is continuous at $x = 0$. If $f^{\prime}(1) = f(-1)$,then the value of $g(3)$ is

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