$f(x)=\frac{e^x-1}{x}$ and $f(0)=1$. Finding $f''(0)$ rapidly.

Hint: If you can use Taylor Series, you could find the series: $$ \begin{align} \frac{e^x-1}{x} &=\frac{\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\dots\right)-1}{x}\\ &=1+\frac{x}{2}+\frac{x^2}{6}+\frac{x^3}{24}+\dots \end{align} $$ and take two derivatives at $x=0$.