# Taylor series with initial value of infinity

So here's the problem:

Let $$f(x)=(1+3x)\cdot e^{\frac{1}{x-2}}$$ Find $a$, $b$ and $c$ such that: $$f(x)=ax+b+\frac{c}{x}+o(1/x)$$ as $x\to \infty$.

I guess what it's asking of me is to find a Taylor series at initial point of $x$ = infinity. In my textbox the problem is solved by first finding the series of the function

and then just multiplying it with the other part which is a polynomial itself. That's all good.

However, two things I don't understand here.

First: How do I deal with orders of less than 1:

How does that even go in the Taylor series formula?

Second: How do I deal with the initial point being infinity? I mean I can see that:

But what about the first derivative, for example:

In order to plug in into the Taylor formula I would need to evaluate it at infinity, which seems rather difficult so I guess I'm doing something wrong here.

Any ideas?

Thanks

Hint. Note that as $x\to \infty$, $$\frac{1}{x-2}=\frac{1}{x}\cdot\frac{1}{1-2/x}=\frac{1}{x}+\frac{2}{x^2}+o(1/x^2).$$ Now use the Taylor expansion $e^t=1+t+\frac{t^2}{2}+o(t^2)$ at $0$: for $t=\frac{1}{x}+\frac{2}{x^2}+o(1/x^2)\to 0$, \begin{align}\exp\left(\frac{1}{x-2}\right)&= 1+\left(\frac{1}{x}+\frac{2}{x^2}+o(1/x^2)\right) +\frac{1}{2}\left(\frac{1}{x}+o(1/x)\right)^2+o(1/x^2)\\ &=1+\frac{1}{x}+\frac{2}{x^2} +\frac{1}{2x^2}+o(1/x^2)=1+\frac{1}{x}+\frac{5}{2x^2} +o(1/x^2).\end{align} Can you take it from here?