= Solution
{c}
Let $g_x(t)=(1+xt)^{-1}$. Repeated <integration by parts> using the factor $e^{-t}$ gives
$$
I(x)=\sum_{n=0}^Ng_x^{(n)}(0)+\int_0^\infty e^{-t}g_x^{(N+1)}(t)dt.
$$
Boundary terms at infinity vanish, and $g_x^{(n)}(t)=(-1)^nn!x^n(1+xt)^{-n-1}$. Therefore
$$
I(x)=\sum_{n=0}^N(-1)^nn!x^n+R_N(x),\qquad
|R_N(x)|\leq(N+1)!x^{N+1}\quad(x>0).
$$
For every fixed $N$, this is $o(x^N)$ as $x\downarrow0$, proving
$$
\boxed{I(x)\sim\sum_{n\geq0}(-1)^nn!x^n.}
$$
For each fixed $x>0$ the displayed infinite series diverges, since successive term magnitudes have ratio $(n+1)x\to\infty$. Its usefulness is precisely that of an <asymptotic expansion>, not a convergent power series.
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