Resilient L'Hospital's rule question I'm trying to show that
$$\lim_{x\to 0} \frac{(1+x)^{\frac{1}{x}}-e}{x} = -\frac{e}{2}$$
At first it seemed like a routine application of L'Hospital's rule, but my standard bag of tricks isn't working. The $e$ in the numerator prevents any log trickery from separating nicely, and the limit being negative seems to also preclude analyzing the limit of the log. 
I tried to interpret this as the derivative of some function at a point, say, $g(u) = u^{\frac{1}{u-1}}$ and the point $u = 1$, but evaluating $g'(1)$ just got worse, and I had concerns about differentiability of $g$ there. Would choosing a different function work out better?
I tried fiddling with one of the limit definitions for $e$ because the first term in the numerator tends to $e$ as $x\to 0$, but the function we're taking the limit of is not continuous at $x=0$ and so moving the limit in was a no-go.
Edit: the $e$ in the numerator seems critical, as the limit diverges without it.
I have a feeling I'm missing something simple. Any hints/solutions would be appreciated.

Hint:
$$(1+x)^{1/x}=e^{\frac{1}{x}\log(1+x)}\xrightarrow [x\to 0]{}e^1=e$$
and thus we can apply L'H:
$$\left[(1+x)^{1/x}\right]'=\left(e^{\frac{1}{x}\log(1+x)}\right)'=\left[-\frac{1}{x^2}\log(1+x)+\frac{1}{x+x^2}\right]e^{\frac{1}{x}\log(1+x)}\ldots$$
