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Solution

Let $\ln L = 2x \ln\left(1 + \frac{a}{x} + \frac{b}{x^2}\right)$ Using expansion $\ln(1+z) = z - \frac{z^2}{2} + \dots$ $\ln L = 2x\left(\frac{a}{x} + \frac{b}{x^2} - \frac{a^2}{2x^2}\right)$ $= 2a + \frac{2(b - a^2/2)}{x} + \dots$ As $x \to \infty$, $\ln L \to 2a$ Given $\ln L = 2 \Rightarrow 2a = 2 \Rightarrow a = 1$. Hence, $b$ can be any real number.