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Find a function $g(x)$ that is as simple as possible s.t. $\exp(x)-1=\mathcal{O}(g(x))$ for $x\to 0$.

Claim. Such a possible function is $g(x)=x$.

Proof. Using the definition of the class $\mathcal{O}$ via limits and the limit of the difference quotient we get the following

$$\lim_{x\to 0}\frac{\exp(x)-1}{x} = \lim_{x\to 0}\frac{\exp(0 + x)-\exp(0)}{x} = (\exp(0))' \overset{(*)}{=} \exp(0) = 1<\infty.$$

Hence $\exp(x)-1=\mathcal{O}(x).$ $\tag*{$\blacksquare$}$

I was curious now as the evaluation of $\exp(0)$ to $1$ at $(*)$ and deriving it afterwards would yield $0$ instead of $1$ even though both values still would be less than infinity. I was wondering whether there are just multiple alternatives or the actual evaluation would be just wrong in this particular case. What's the actual reasoning in this case?

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You can just apply L'Hopital's rule, since the limit of of the form $\frac{0}{0}$: $$\lim_{x \to 0}{\frac{e^x-1}{x}}=\lim_{x \to 0}{\frac{e^x}{1}=1.}$$

EDIT: In your answer, you always differentiate first before you evaluate the function at a point, and so obtain $\displaystyle \left.e^x\right|_{x=0}=e^0=1$ rather than $(e^0)'=1'=0.$

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