Well, I've tried using logarithms (a special case of L'Hopital's rule, actually), like this:

$\displaystyle e=\lim_{x->\infty}\left(1+\frac{1+x}{4x^2+4x+4}\right)^{4x+1 }=\lim_{x->\infty}\left(1+4x)\log_{e}\left(1+\frac{1+x}{4x^2 +4x+4}\right)$

But then we have again a case of $\displaystyle 0*\infty$, and another application of L'Hopital's rule is needed, and then another and... then, some derivatives need to be calculated and finally, when all is done, I get the right result e.

I have a hunch there exists a more elegant path to this result, but unfortunately I don't see it at the moment. Care to help?