Find the highest $n\in \mathbb N$ such that $f^{(n)}$ exists 
Let $f: \mathbb R\rightarrow \mathbb R$ which for $x \in \mathbb R$: $$f(x)=\begin{cases}\frac{e^x-1}{x}, & x\neq 0\\1, & x=0\end{cases}$$ Find $\sup\left\{ n\in \mathbb N: f\text{ is }n\text{-times differentiable}\right\} $.

My try:For starters, let's check if $f$ has $f^{(1)}$:For $x\neq 0$: 


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*$f$ is a composite of elemental differentiable functions so also $f$ is differentiable for $x\neq 0$
For $x=0$: 
$$f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\rightarrow 0}\frac{\frac{e^x-1}{x}-1}{x}=\lim_{x\rightarrow 0}\frac{e^x-1-x}{x^2}=\lim_{x\rightarrow 0}\frac{1+x+\frac{x^2}{2}+o(x^2)-1-x}{x^2}=\frac{1}{2}+\frac{o(x^2)}{x^2}=\frac{1}{2}$$However we know that for $x=0$: $f(x)=1$ so $f'(x)=0\neq \frac{1}{2}$That's why in my opinion $f$ hasn't $f'$ so also other derivative not exist.  However I think I can have an error because then this task is really easy.Can you check it? 
 A: If we know that $f(x)=1$ for all $x$ in an interval around $0,$ then we can conclude that $f'(0)=0,$ but this is not the case. Let me apply the same argument to $g(x)=x$ to show you the issue.

"Let $\alpha$ be any real number. On the one hand, $$g'(\alpha):=\lim_{x\to \alpha}\frac{g(x)-g(\alpha)}{x-\alpha}=\lim_{x\to \alpha}\frac{x-\alpha}{x-\alpha}=\lim_{x\to \alpha}1=1.$$ But on the other, we know that $g(\alpha)=\alpha,$ so since $\alpha'=0,$ then $g'(\alpha)=0\ne 1.$ Thus, $f$ is nowhere differentiable."

Rather, you have directly calculated $f'(0)=\frac12.$ Thus, $f$ is once differentiable, at least.

It turns out that we can do even better, using the fact that $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!},$$ so that $$e^x-1=\sum_{n=1}^\infty\frac{x^n}{n!}=x\sum_{n=1}^\infty\frac{x^{n-1}}{n!}=x\sum_{n=0}^\infty\frac{x^n}{(n+1)!}.$$
At this point, we can fairly directly prove that $$f(x)=\sum_{n=0}^\infty\frac{x^n}{(n+1)!}$$ for all $x\in\Bbb R.$ Thus, $f$ is differentiable as many times as we like at $0,$ and the $n$th derivative will be $\frac1{n+1}$ for all $n.$ Since $f$ is differentiable as many times as we like everywhere else, too, then $f$ is differentiable as many times as we like everywhere.
A: $$\frac{e^x- 1}x = \frac{\sum_{k=0}^\infty \frac{x^{k}}{k!}-1}x = \frac{\sum_{k=1}^\infty \frac{x^{k}}{k!}}x = \sum_{k=1}^\infty \frac{x^{k-1}}{k!}  =\sum_{n=0}^\infty \frac{x^n}{(n+1)!}=1+\frac{x}2+\frac{x^2}{6}+\dots$$
So this function is analytic with convergent everywhere power series $f(x)=\sum_{n=0}^\infty a_n x^n$ given above. The derivatives at $0$ are $$f^{(n)}(0) = n!a_n = \frac{1}{n+1}.$$
I have no idea what you mean by 

$\text { However we know that for } x=0 : f(x)=1 \text { so } f^{\prime}(x)=0 \neq \frac{1}{2}$

Graph of $f$, some derivatives and a 5th order Taylor polynomial:

This function actually has a name, it is the  (1,2)-Mittag-Leffler function
$$ f(x) = E_{1,2}(x) = \sum_{n=0}^\infty \frac{x^n}{\Gamma(n+2)}.$$
For similar reasons, $\operatorname{sinc}(x):=\frac{\sin(x)}x$ only has a removable singularity at 0.
