If $f(x)=\frac{e^x}{x}$, evaluate $\lim\limits_{n\to\infty}\frac{f^{(n)}(1)}{n!}$. 
If $f(x)=\frac{e^x}{x}$, evaluate $$\lim_{n\to\infty}\frac{f^{(n)}(1)}{n!}$$ where $f^{(n)}(x)$ is the $n$-th derivative of $f(x)$.

This is from a local contest held in my city last year. I tried by calculating first few derivatives of $f(x)$ to have an estimation of $f^{(n)}(x)$ and got the following results: $$\begin{align} f'(x) &= \frac{e^x(x-1)}{x^2}\\ f''(x) &= \frac{e^x(x^2-2x+2)}{x^3}\\ f'''(x) &= \frac{e^x(x^3-3x^2+6x-6)}{x^4}
\end{align}$$
As it can be seen, the derivatives are getting complicated in each step and I can't expect to have a nice expression for $f^{(n)}(x)$. Then how do I solve the problem?
Also, I would like to share my current knowledge of the topic to receive an answer that I can understand. I'm currently a high schooler and am learning calculus as an amateur. I have done almost all elementary concepts of differentiation and limits. I have not learnt integration yet. I don't know about Taylor and Maclaurin series but I know the expansions of $e^x$, $\ln(1+x)$ etc. I hope the answerers will consider these and provide a detailed solution if needed.
Edit: As figured out in the comments, the limit should be $$\lim_{n\to\infty}\bigg|\frac{f^{(n)}(1)}{n!}\bigg|.$$
Sorry for the inconveniences.
 A: Equivalently, define $g(x) = \frac{e^x}{x + 1}$. Then we seek $e\lim\limits_{n \to \infty} \frac{g^{(n)}(0)}{n!}$.
Now consider that the MacLaurin series for $e^x$ is $\sum\limits_{i = 0}^\infty \frac{x^i}{i!}$. And consider that the MacLaurin series for $\frac{1}{1 + x}$ is $\sum\limits_{i = 0}^\infty (-1)^i x^i$.
Therefore the MacLaurin series for $g(x)$ is $\sum\limits_{i = 0}^\infty x^i \sum\limits_{j = 0}^i \frac{1}{j!} (-1)^{i - j}$. The idea is that when we multiply two MacLaurin series, we can collect all the terms of the form $c x^i$ just like when we multiply two polynomials.
From this, we see that $\frac{g^{(n)}(0)}{n!} = \sum\limits_{j = 0}^n \frac{1}{j!} (-1)^{n - j} = (-1)^n \sum\limits_{i = 0}^n \frac{(-1)^j}{j!}$.
Therefore, we see that $\lim\limits_{n \to \infty} \frac{g^{(n)}(0)}{n!}$ does not exist. For note that $\lim\limits_{n \to \infty} \sum\limits_{i = 0}^n \frac{(-1)^j}{j!} = e^{-1}$. Therefore, for sufficiently large $n$, we have $\frac{g^{(n)}}{n!}$ is very close to $e^{-1}$ when $n$ is even, and very close to $-e^{-1}$ when $n$ is odd. This proves that the limit does not exist.
If you throw in an absolute value, then we see that $\lim\limits_{n \to \infty} | \frac{g^{(n)}(0)}{n!}| = e^{-1}$ and thus $\lim\limits_{n \to \infty} |\frac{f^{(n)}(1)}{n!}| = 1$.
A: I think the question should be
$$ \lim_{n\to\infty}\bigg|\frac{f^{(n)}(1)}{n!}\bigg|=1. $$
Clearly $e^x=xf(x)$. By Leibniz's Rule, one has
$$ (e^x)^{(n)}=(xf(x))^{(n)}=\sum_{k=0}^n\binom{n}{k}x^{(k)}f^{(n-k)}(x)$$
which gives
$$ e^x=xf^{(n)}(x)+nf^{(n-1)}(x). $$
So
$$f^{(n)}(1)=e-nf^{(n-1)}(1)\tag{1} $$
and hence
$$ \frac{f^{(n)}(1)}{n!}=\frac{e}{n!}-\frac{f^{(n-1)}(1)}{(n-1)!}. \tag{2}$$
Continuing to use (1) in (2), one has, for $n=2k$,
\begin{eqnarray} 
\frac{f^{(2k)}(1)}{(2k)!}&=&\frac{e}{(2k)!}-\frac{e}{(2k-1)!}+\frac{f^{(2k-2)}(1)}{(2k-2)!}\\
&=&\cdots\\
&=&e\sum_{i=1}^{k}\bigg(\frac{1}{(2i)!}-\frac{1}{(2i-1)!}\bigg)+e,\\
&=&e\bigg(1-\frac{1}{1!}+\frac{1}{2!}-\cdots-\frac{1}{(2k-1}+\frac{1}{(2k)!}\bigg)\\
&\to&e\cdot e^{-1}=1, \text{as }k\to\infty.
\end{eqnarray}
For $n=2k-1$,
\begin{eqnarray} 
\frac{f^{(2k-1)}(1)}{(2k-1)!}&=&\frac{e}{(2k-1)!}-\frac{f^{(2k-2)}(1)}{(2k-2)!}\\
&\to&-1, \text{as }k\to\infty.
\end{eqnarray}
So
$$ \lim_{n\to\infty}\bigg|\frac{f^{(n)}(1)}{n!}\bigg|=1. $$
A: Notice that
$$f(x):=\frac{e^x}{x}=e\frac{1}{1+(x-1)}e^{x-1}=e\Big(\sum^\infty_{n=0}(-1)^n(x-1)^n\Big)\Big(\sum^\infty_{n=0}\frac{(x-1)^n}{n!}\Big)$$
By Cauchy's product formula and Taylor's theorem the $n$-th coefficient of the power series about $x=1$ above is
$$\frac{f^{(n)}(1)}{n!}=c_n=e\sum^n_{k=0}\frac{(-1)^{n-k}}{k!}=(-1)^ne\sum^n_{k=0}\frac{(-1)^k}{k!}$$
$$(-1)^n\frac{f^{(n)}(1)}{n!}\xrightarrow{n\rightarrow\infty}1$$
since $\sum^n_{k=0}\frac{(-1)^k}{k!}\xrightarrow{n\rightarrow\infty}e^{-1}$.
A: We will prove by induction that
$$f^{(n)}(x)=\frac{e^x}{x^{n+1}}P_n(x)$$
where $P_n(x)$ is a polynomial with the form
$$P_n(x)=\sum_{i=0}^n\frac{n!}{i!}(-1)^{i+n}x^i$$
The first few of these polynomials are
$$P_0(x)=1$$
$$P_1(x)=x-1$$
$$P_2(x)=x^2-2x+2$$
$$\vdots$$
Since the proposition is obvious for the base case, we proceed to the induction step. Assume that
$$f^{(n)}(x)=\frac{e^x}{x^{n+1}}P_n(x)$$
Then
$$f^{(n+1)}(x)=\frac{e^x}{x^{n+2}}  \left((-n+x-1) P_n(x)+x P_n^{'}(x)\right)$$
Thus, if we can prove $A=(-n+x-1) P_n(x)+x P_n^{'}(x)-P_{n+1}(x)=0$ we are done. We have that
$$A=(-n+x-1) P_n(x)+x P_n^{'}(x)-P_{n+1}(x)$$
$$=(-n+x-1)\sum_{i=0}^n\frac{n!}{i!}(-1)^{i+n}x^i+x\sum_{i=1}^n\frac{n!}{(i-1)!}(-1)^{i+n}x^{i-1}-\sum_{i=0}^{n+1}\frac{(n+1)!}{i!}(-1)^{i+n+1}x^i$$
Now, lets look at an arbitrary $i\in\{1,2,...,n\}$. This term is
$$(-n+x-1)\frac{n!}{i!}(-1)^{i+n}x^i+\frac{n!}{(i-1)!}(-1)^{i+n}x^{i}-\frac{(n+1)!}{i!}(-1)^{i+n+1}x^i$$
Collecting like terms in a coefficient gives us
$$=\frac{n!}{(i-1)!}(-1)^{i+n}x^i\left(\frac{x}{i}+1\right)$$
Putting this back into the equation above for $A$ gives us
$$A=(-n+x-1)n!(-1)^n+(n+1)!(-1)^n-x^{n+1}+\sum_{i=1}^n\frac{n!}{(i-1)!}(-1)^{i+n}x^i\left(\frac{x}{i}+1\right)$$
$$=n!(-1)^n\left[x+\sum_{i=1}^n\frac{1}{(i-1)!}(-1)^{i}x^i\left(\frac{x}{i}+1\right)-(-1)^nx^{n+1}\right]$$
This inner expression can then be simplified to
$$x+\sum_{i=1}^n\frac{1}{(i-1)!}(-1)^{i}x^i\left(\frac{x}{i}+1\right)-(-1)^nx^{n+1}$$
$$=x+\sum_{i=1}^n\frac{1}{(i-1)!}(-1)^{i}x^{i}+\sum_{i=1}^n\frac{1}{i!}(-1)^{i}x^{i+1}-(-1)^nx^{n+1}$$
$$=+\sum_{i=1}^{n-1}\frac{1}{i!}(-1)^{i+1}x^{i+1}+\sum_{i=1}^{n-1}\frac{1}{i!}(-1)^{i}x^{i+1}=0$$
as desired. Thus, we know that
$$f^{(n)}(x)=\frac{e^x}{x^{n+1}}\sum_{i=0}^n\frac{n!}{i!}(-1)^{i+n}x^i$$
$$f^{(n)}(1)=e\sum_{i=0}^n\frac{n!}{i!}(-1)^{i+n}$$
This implies that
$$\lim_{n\to\infty}\frac{f^{(n)}(1)}{n!}=e\lim_{n\to\infty}\sum_{i=0}^n\frac{1}{i!}(-1)^{i+n}=e\lim_{n\to\infty}(-1)^n\sum_{i=0}^n\frac{1}{i!}(-1)^{i}$$
This limit does not exist since the sum converges to $e^{-1}$ while $(-1)^n$ oscillates. However, if we take the absolute value we get
$$\lim_{n\to\infty}\left|\frac{f^{(n)}(1)}{n!}\right|=\lim_{n\to\infty}\left|e\sum_{i=0}^n\frac{1}{i!}(-1)^{i}\right|=ee^{-1}=1$$
A: Note that $x \mapsto f(x)-\frac{1}{x}$ is an entire function (by defining it to be $1$ at $x=0$). Whence the limit of the absolute values of the Taylor coefficients about $x=1$ will agree with that of the function $\frac{1}{x}$, which is just $1$ since
$$
\frac{1}{x} = \frac{1}{{1 - (1 - x)}} = \sum\limits_{n = 0}^\infty  {(1 - x)^n }  = \sum\limits_{n = 0}^\infty  {( - 1)^n (x - 1)^n } .
$$
