Maclaurin expansion series at $x=0$

Question

I have the function $$f(x)=\left\{\begin{array}{cl} \dfrac{e^{x}-1}{x}, & x\neq0 \\ 1, & x=0 \end{array}\right.$$ And I gotta find its Maclaurin expansion series.

I know how to find the one from $\dfrac{e^{x}-1}{x}$. But how do get a function that $f(0)=1$?

Answer 1

$$e^x - 1 = \sum_{k\ge 1} {x^k\over k!}.$$ Now divide by $x$ and you are done.

Answer 2

Your constant is equal to 1, as your function approaches 1 when x=0. So set your constant to 1 and just use derivatives to approximate $\dfrac{e^{x}-1}{x}$ around x=0.