How can I show $ \lim_{x\to -\infty}\frac{1}{(e^x)(x-1)}=-\infty$? I need to find this limit:
$$ \lim_{x\to -\infty}\frac{1}{(e^x)(x-1)}$$
Now I am given the answer as $-\infty$, and that makes some sense as I use series expansion. But I was thinking of any other ways to find this limit rather than series?
I know for one that I can check the highest degree of $x$ in the denominator and divide above and below by that value. However, here I am confused. Should I divide throughout by $e^x$ or $x$?
 A: Maybe switch the argument to
$$ 
\lim_{x\to -\infty} \frac{e^{-x}}{x-1}
$$
And then use L'Hospital's Rule after confirming an indeterminate form?
A: First we put $y = -x$ or $x = -y$.
Then the given limit
$$
\tilde{L} = \lim\limits_{y\rightarrow \infty} \
{1 \over e^{-y} \ (-y-1)}
= - \lim\limits_{y \rightarrow \infty} \ {e^y \over 1 + y} = - L
\tag{1}
$$
where
$$
L = \lim\limits_{y \rightarrow \infty} \ {e^y \over 1 + y} \tag{2}
$$
We put $z = 1 + y$. Then (2) becomes
$$
L = \lim\limits_{z \rightarrow \infty} \ {e^{z-1} \over z} =
\lim\limits_{z \rightarrow \infty} \ {e^{z} e^{-1} \over z}
= {1 \over e} \lim\limits_{z \rightarrow \infty} \ {e^{z}  \over z}
\tag{3}
$$
We can simplify (3) as
$$
L = {1 \over e} \ \lim\limits_{z \rightarrow \infty} \
\left[ {1 + z + {z^2 \over 2!} + {z^3 \over 3!} + \cdots 
\over z } \right] 
$$
Thus,
$$
L = 
{1 \over e} \ \lim\limits_{z \rightarrow \infty} \
\left[ {1 \over z} + 1 + {z \over 2!} + {z^2 \over 3!} + \cdots
\right] = \infty
$$
Substituting $L = \infty$ into (1), we deduce that
$$
\tilde{L} = -\infty
$$
