\begin{align} \lim_{t \to 0} f(x,t) &= \lim_{t \to 0} \frac{t \left( 1 + xt + \cdots \right)}{t + \frac{1}{2} t^2 + \cdots} \\ &= \lim_{t \to 0} \frac{1 + xt + \cdots}{1 + \frac{1}{2} t + \cdots} \\ &= 1 \end{align}
\begin{align} g(t) &= \frac{t \left( e^t + 1 \right)}{2 \left( e^t - 1 \right)} \\ g(-t) &= \frac{-t \left( e^{-t} + 1 \right)}{2 \left( e^{-t} - 1 \right)} \\ &= \frac{-t \left( 1 + e^t \right)}{2 \left( 1 - e^t \right)} \\ &= \frac{t \left( e^t + 1 \right)}{2 \left( e^t - 1 \right)} \\ &= g(t) \end{align} なので、 $g(t)$ は偶関数である。
\begin{align} \frac{\partial f(x,t)}{\partial x} &= \frac{t^2 e^{xt}}{e^t - 1} \\ \frac{\partial f(x,t)}{\partial t} &= \frac{\left( e^{xt} + xt e^{xt} \right) \left( e^t - 1 \right) - t e^{xt} e^t } {\left( e^t - 1 \right)^2} \\ &= \frac{e^{xt} \left( e^t - 1 + xte^t - xt - te^t \right)}{\left( e^t - 1 \right)^2} \end{align}
\begin{align} \lim_{t \to 0} \frac{\partial f(x,t)}{\partial t} &= \lim_{t \to 0} \frac{e^{xt} \left( e^t - 1 + xte^t - xt - te^t \right)}{\left( e^t - 1 \right)^2} \\ &= \lim_{t \to 0} \frac{\left( x - \frac{1}{2} \right) t^2 + \cdots }{t^2 + \cdots} \\ &= x - \frac{1}{2} \end{align}