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 $n_0$). 
 A: 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 $\infty$.
A: From the power series we can obtain $f'(x) = -f(x)$. Also $f(0) = 1$. Using these we will show that $f(x) > 0$ for all $x$.
Let's suppose that there is a number $a$ such that $f(a) = 0$ and consider the function $$g(x) = f(a - x)f(x - a)$$ then its derivative is given by
\begin{align}
g'(x) &= -f'(a - x)f(x - a) + f(a - x)f'(x - a)\notag\\
&= f(a - x)f(x - a) - f(a - x)f(x - a)\notag\\
&= 0\notag
\end{align}
so that $g(x)$ is a constant and therefore $1 = g(a) = g(0) = 0$ and we get a contradiction. Hence $f(x) \neq 0$ for all $x$. Since $f(0) = 1$ it follows by continuity that that $f(x) > 0$ for all $x$.
From $f'(x) = -f(x) < 0$ we see that $f(x)$ is strictly decreasing and since it is bounded below by $0$, the limit $\lim_{x \to \infty}f(x) = L$ exists. Hence $\lim_{x \to \infty}f'(x) = -L$ also exists. Now by mean value theorem we have $$f(x + 1) - f(x) = f'(\xi)\tag{1}$$ where $x < \xi < x + 1$. Taking limit as $x \to \infty$ on both sides of $(1)$ we get $L - L = -L$ so that $L = 0$. Hence $\lim_{x \to \infty}f(x) = 0$.
Note: As noted by PhoemueX in his answer, there is no general theorem to prove that a power series tends to $0$ as $x \to \infty$ based on certain criteria for its coefficients. Each power series needs to be analyzed properly in order to determine its behavior near $\pm\infty$. Here I managed to get a nice property of $f(x)$ namely $f'(x) = -f(x)$ from the series representation which is very useful in the analysis given in my answer.
A: One approach would be to show, via the power series than $f(N) = f(1)^N$ focusing on integer $N > 1$ so your power series manipulations are manageable.  Then establish that $f(1) < 1$.  It's easy from here to show that $f(N)$ converges to $0$ in the limit of large $N$.  Now you must prove that $f(x) \leq f([x])$ for any real $x$ where $[x]$ is the greatest integer not exceeding $x$.
A: I think the correct strategy in other situations is to look for something analogous to $\exp(t)\exp(-t)=1$.
A: Another obstruction is that the proof has to work with $x$ replaced with $ax$ for positive $a$, which changes the $n$'th coefficient by a factor of $a^n$.
A more accessible question is to find nice conditions on coefficients of a formal power series $1 + \sum a_n x^n$ so that the logarithm or the inverse of the series have negative coefficients.
