Is $e^{|x|}$ differentiable?

NB: The limit I'm referring to isn't $\lim_{x\to 0^{\pm}} f'(x)$ as some commenters thought. Rather, it's $\lim_{h\to 0^{\pm}}\frac{f(0+h)-f(0)}{h}$

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Your logic at the end is correct. If the function were to be differentiable, then necessarily the left and right limits exist and agree, so we can check that. The left-hand limit is the derivative of $e^{-x}$ at $0$ and the right hand limit is the derivative of $e^x$ at $0$. At zero the former is $-1$ and the later is $1$ so the limit doesn't exist, and the function isn't differentiable.

You have a decent idea; however, the derivative of a function is not guaranteed to be continuous, so the argument doesn't work as is. Here is a way to fix it.

By Darboux's theorem, if $g$ is the derivative of a function, then $g$ has the intermediate value property.

Away from zero, the derivative of $f(x) = e^{|x|}$ is given by

$$ f'(x) = \operatorname{sign}(x) e^{|x|} $$

It's fairly easy to see that this function doesn't attain any values in the interval $[-1, 1]$ for nonzero $x$.

However, since $f$ does have values larger than $1$ and smaller than $-1$, the intermediate value property implies every value in $[-1, 1]$ has to be attained someplace.

But the only possible place is at $x=0$, and $f'(0)$ cannot be every number in $[-1,1]$ simultaneously!

Thus, $f$ is not differentiable.

Yes, you are right. Note that the function $y=e^{|x|}$ is symmetric with respect to the $y$axis so at $x=0$ we have a cusp because the tangent of $e^x$ at this point is not null.