This has been bugging me for hours now. The question as it stands is;
Let $$f(x) = {1 \over x} - {1 \over e^x-1}$$ for all $x \neq 0$
By writing $f(x)$ as the quotient of 2 power series, find $\lim_{x\to 0} f(x)$.
Now I realise $e^x = 1 + x + {x^2 \over 2!} + \cdots $ and that I can rearrange $f(x) = {e^x -1 -x \over x(e^x -1)} $ and in doing so can now write it using the expansion as;
$$ {x^2 /2! + x^3 /3! + \cdots \over x^2 + x^3 /2! + x^4 /3! + \cdots}$$
And clearly then I can get it to;
$$ {1 /2! + x /3! + ... \over 1 + x /2! + x^2 /3! + ...} $$
And then obviously this $\to {1/2 \over 1} = 1/2$ as $x \to 0$.
My issue here is no matter where I look the series expansion for $f(x)$ contains alternate terms, positive and negative, and my expansion is therefore incorrect. I know however that the limit is correct but I have no idea how to deduce the correct expansion without using the taylor series. In these questions we are not told (nor have we used) taylor series etc, however we haven't been told to not use it. I am assuming there is another way hence why we haven't been directly told to use it.
The second part to the question is asking me to compare the series term by term in order to show that $f(x) \le 1/2$ for all $x>0$ which is clear from the correct expansion.
Many thanks for any help!!
EDIT: Following the comment below I have noticed in that case that the expansion with alternate terms is indeed a taylor series. So I guess my question now is if the expansion I have wrote above would be correct?