Does the function $\dfrac{\cos(x)}{x}$ have a Taylor expansion at $x=0$

Question

I am doing some formal manipulation of series right now and stumble across this function $\dfrac{\cos(x)}{x}$

The series expansion that I carry it out is: $\dfrac{\cos(x)}{x}=\dfrac{1}{x}-\dfrac{x}{2!}+\dfrac{x^3}{4!}-\dfrac{x^5}{6!}...$

The limit as $x$ approaches $0$ from the left is$-\infty$, while the limit as $x$ approaches from right hand is $+\infty$. So the limit doesn't exist. Does this mean the function do not have a Taylor expansion?

I look up Wolfram and it says that this function has a series expansion, why?0

Answer 1

This is not a Taylor expansion because of that $1/x$ term. Taylor series can only have nonnegative integer powers. Instead, it is a Laurent series.