Limit a compound function

First, we find the power series expansion of $e^x$ and $1/x$ around $x=1$ : $e^x=e\cdot e^{x-1}=\sum_{k=0}^\infty\frac e{k!}(x-1)^k$ $\frac1x=\frac1{x-1+1}=\sum_{k=0}^\infty(-1)^k(x-1)^k$ $\implies \frac{e^x}x=\left(\sum_{k=0}^\infty\frac e{k!}(x-1)^k\right)\left(\sum_{k=0}^\infty(-1)^k(x-1)^k\right)$ If we expand this product out, the coefficient of $(x-1)^n$ is given by $\sum_{k=0}^n(-1)^{n-k}\frac e{k!}$ The expanded polynomial is called the Cauchy product , which converges around $x=1$ since the two series both converge absolutely around $x=1$ .

If we take the derivative $2n$ times and evaluate at $x=1$ , we get the coefficient of the $(x-1)^{2n}$ term times $(2n)!$ , hence $\frac{f^{(2n)}(1)}{(2n)!}=\sum_{k=0}^{2n}(-1)^{2n-k}\frac e{k!}=e\sum_{k=0}^{2n}\frac{(-1)^k}{k!}$ As $n$ approaches infinity, this sum approaches $e^{-1}$ , so the result of the limit is $e\times e^{-1}=1$ .