Limit of $\lim_{x \to \infty} f(x)$ using power series of $f$ Suppose we want to find $\lim_{x \to \infty} f(x)$. 
Moreover, we know that $f(x)$ has a power series expension for all $x \in \mathbb{R}$ 
\begin{align}
f(x)=\sum_{n=0}^\infty a_n x^n
\end{align}
and we have a nice closed expression for all $a_n$'s. 
My question can we compute $\lim_{x \to \infty} f(x)$ based on the knowledge of $a_n$'s?  
Case 1: If $a_n \ge 0$ then 
\begin{align}
f(x) \ge a_0+a_1x
\end{align}
so
\begin{align}
 \lim_{ n\to \infty} f(x)= \infty. 
\end{align}
Case 2:  An interesting case which I would like to understand is when $a_n$'s have alternating sign and $|a_{n+1}| \ge |a_n|$.
Thanks for any help. 

It seems like a very delicate question; take for example $f(x)=\sin(x)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}x^{2k+1}$ and $g(x)=\frac{\sin(x)}{x}=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}x^{2k}$. Then $f$ ad $g$ have both infinite radius of convergence and very similar coefficients, but $\lim_{x\to\infty}f(x)$ doesn't exist, while $\lim_{x\to\infty}g(x)=0$. So it seems that its difficult to directly tell something about the limit solely by considering the sequence of coefficients, without calculating $f$ (and thereby the limit) explicitly.
