How to show that $e^{-x}$ tends to $0$ when $x\to \infty$ if $e^{-x}$ is defined as the power series. With only the formal definition of $$f(x) = \exp(-x)= \sum \frac{(-x)^n}{n!}$$ how can we show that $$\lim_{x\to  \infty} f(x)=0?$$
I am looking for a proof that would not use the identity $\exp(x)\exp(-x)=1$ in order to find a strategy for similar problems (i.e. limits of functions defined as power series $\sum a_n x^n$ where $a_n$ does not have a fixed sign beginning at any $n_0$). 

I don't think that this is possible.
The proof has to be very specific to the given series because of the following: Consider $g(x) = \sum a_n x^n$ with $a_n = \frac{1}{n!}$ for all $n \neq n_0$ and $a_n = \frac{1}{n!} + \delta$ for some $\delta > 0$ and some fixed even integer $n_0 \geq 2$.
We then have
$$
g(x) = e^x + \delta x^{n_0} \xrightarrow[x \to -\infty]{} \infty,
$$
since the first summand vanishes and the second one goes to $\infty$.
This shows that just chaning a single coefficient with an arbitrarily small perturbation makes the claim invalid.
BTW: The same argument works for odd $n_0$, but in this case, we get convergence to $-\infty$ instead of $\infty$.
