About L'Hôpital's Rule applied to some limits and uniform convergence

About L'Hôpital's Rule applied to some limits and uniform convergence

I just wanted to check whether I was missing something.
An exercise/ exampl ein my text shows the following: $$f_n(x) =
nxe^{-nx};\ 0 \le x \le 1$$
and says to see if it converges uniformly you can apply L'Hôpital's rule.
Since when x=0 the whole thing converges to zero we can check using that
if it does the same thing elsewhere on the interval. When I do that I get:
$$\lim_{n \rightarrow \infty} f_n(x) =
\frac{\frac{d}{dx}(nx)}{\frac{d}{dx} (e^{nx})}= \frac{n}{ne^{nx}}=
\frac{1}{e^{nx}}$$ and that converges to zero except at $f(\frac{1}{n})$
which is $e^{-1}$ and does not converge to zero. In addition when you plug
in zero for x you get $\frac{1}{e^0} = 1$ SO we know this isn't uniformly
converging.
But here's where it gets odd (for me). The book has this solution:
$$\lim_{n \rightarrow \infty} | f_n(x) |= |x| \lim_{n \rightarrow \infty}
\frac{n}{e^{-nx}} = |x| \lim_{n \rightarrow \infty}
\frac{1}{|x|e^{-n}}=0$$
ANd I am a bit confused. First, wouldn't the power of $e$ be positive on
the bottom half of the fraction? And how did the x in the power of e
disappear in the last bit?
Maybe this is some elementary thing I don't know, but it just looked
wrong. (Thanks for the help -- this site has been a life saver). BTW I
tagged this as PDE since it is in a PDE text.
EDITS: fixed the less than to make them less than or equal to in the
original equation.