Problem in evaluating this limit using Taylor series If one says:
$\lim_{x\rightarrow\infty}\frac{e^{2x}-e^{-2x}-4}{5x^{2}}=\lim_{x\rightarrow\infty}\frac{e^{2x}}{5x^{2}}=\lim_{x\rightarrow\infty}\frac{1+2x+\frac{4x^{2}}{2}+\frac{8x^{3}}{6}+\cdots}{5x^{2}}=\lim_{x\rightarrow\infty}\frac{\frac{8x^{3}}{6}}{5x^{2}}=\lim_{x\rightarrow\infty}\frac{8x^{3}}{30x^{2}}=\infty$
The answer to the limit is correct, but is there a mathematical problem to use Taylor series at 0 as $x\rightarrow\infty$?
 A: You are lucky to have a series with positive coefficients, considered for positive values of $x$. This leads to a simple lower bound for $f$: just drop  all terms except one. The best part is that you get to choose which term to keep.  Let's state the result in greater generality: 
Claim. Suppose that $f:\mathbb R \to \mathbb R $ is a function whose Taylor series at some point $a\in\mathbb R$ converges on $\mathbb R$ and has nonnegative coefficients. Then $f$ is either a polynomial, or it grows at $+\infty$ faster than any polynomial: 
$$\lim_{x\to+\infty} \frac{f(x)}{x^n} = \infty \qquad \forall n\in\mathbb N \tag{1}$$
Proof.   Suppose $f$ is not a polynomial. Given $n$, choose $m>n$ such that the $m$th degree term in the Taylor series of $f$ is nonzero, i.e., $c_m>0$. By the nonnegativity assumption, $f(x)\ge c_m(x-a)^m$ for all $x\ge a$. Since
$$\lim_{x\to+\infty} \frac{c_m(x-a)^m}{x^n} = \infty  $$ 
(1) follows. $\quad\Box$

If your series had terms of varying sign, trying to use it as $x\to \infty$ would likely be   painful.
