When working with a limit of e^x such as:
The limit as x approaches infinity of
Can I simply state that it tends towards 1/((e^inf)/inf) and therefore to 0, or is the limit not yet defined since it tends towards inf/inf below 1? Thanks.
I think you cannot use the Maclaurin series for $e^x$ which is about $x=0$ to analyze $e^x$ when $x\to\infty$. The expansion point (zero) is too far away from the limit point (infinity).
To solve the problem, you have to use L'Hospital's or just know that exponentials grow faster than polynomials.
This statement above is false: $\displaystyle\lim_{x\to\infty}\frac{x}{1-e^x}=-1$. The Maclaurin series approach gives an erroneous answer, because it cannot be used here.
Rather, the limit equals 0, as pointed out in other posts.
$$\begin{aligned}
0 &\lt {1 \over t} \lt t^{c-1} \qquad (t \gt 1,\, c \gt 0) \\
\int_1^x 0 \,\mathrm d t &\lt \int_1^x {1 \over t} \,\mathrm d t \lt \int_1^x t^{c-1} \,\mathrm d t \qquad (x \gt 1) \\
0 &\lt \log x \lt {x^c - 1 \over c} \lt {x^c \over c} \\
0 &\lt \log^a x \lt {x^{ac} \over c^a} \qquad (a \gt 0) \\
0 &\lt {\log^a x \over x^b} \lt {x^{ac-b} \over c^a} \qquad (b \gt 0) \\
\end{aligned}$$
Since this is true for all $c \gt 0$ we may pick any $c = {b \over 2a}$ so that
$$
0 \lt {\log^a x \over x^b} \lt {x^{-{b \over 2}} \over c^a} \\
$$
And now, taking the limit as $x \to \infty$ gives
$$
0 \le \lim_{x \to \infty} {\log^a x \over x^b} \le \lim_{x \to \infty} {x^{-{b \over 2}} \over c^a} = 0 \\ \implies \lim_{x \to \infty} {\log^a x \over x^b} = 0
$$
Then, writing $x = \mathrm e^y$ we have
$$\begin{aligned}
\lim_{x \to \infty} {\log^a x \over x^b} &= 0 \\
\lim_{y \to \infty} {\log^a \mathrm e^y \over \mathrm e^{by}} &= 0 \\
\lim_{y \to \infty} {y^a \over \mathrm e^{by}} &= 0 \\
\end{aligned}$$
Now, addressing your question
$$\begin{aligned}
\lim_{x \to \infty} {x^2 \over 1 - \mathrm e^x} &= \lim_{x \to \infty} {x^2 \mathrm e^{-x} \over \mathrm e^{-x} - 1} \\
&= \lim_{x \to \infty} {{x^2 \over \mathrm e^{x}} \over \mathrm e^{-x} - 1} \\
&= {0 \over 0 - 1} = 0
\end{aligned}$$
Is it x --> + infinity or - infinity. You could alternatively reason this way:
as x --> + infinity (1 - e^x) --> (- e^x)
At the limit the ratio varies as x^2/(-e^x) and the limit is 0 from the left
as x --> - infinity (1 - e^x) --> 1
At the limit the ratio varies as x^2 --> + infinity