Past exam of the mathematical methods II course of the University of Oxford / 2019 resit / 9 / b / Solution

For $|x|<1$, the geometric series gives $(1+x{)}^{-1}={\sum }_{j=0}^{\infty }(-x{)}^j$. Integrating from zero to $x$ yields the Maclaurin seriesOn the real axis the series also converges at $x=1$ to $\ln 2$, but diverges at $x=-1$.

For fixed real $a$ and sufficiently large positive $x$, the base in the requested limit is positive. Apply the series with argument $-a^2/x^2$:Continuity of the exponential function therefore givesThis includes $a=0$, when the expression is identically one.