Behaviour of modified Bessel function of the first kind $I_{\nu}(x)$ As stated in the comments to my previous question, the modified Bessel function of the second kind can be defined as
$$K_{\nu}(x) = \frac{\pi}{2}i^{1 + \nu} (J_{\nu}(ix) + iN_{\nu}(ix))$$
The modified Bessel function of the first kind instead can be defined as
$$I_{\nu}(x) = e^{-i \frac{\pi}{2} \nu} J_{\nu} (ix)$$
So, there is a sort of asymmetry between $I_{\nu}(x)$ and $K_{\nu}(x)$. 
1) Is it normal or can $I_{\nu}(x)$ be expressed as proportional to $J_{\nu}(ix) - iN_{\nu}(ix)$?
This is a problem when trying to obtain the asymptotic behaviour of the function for $x \gg 1$. It should be proportional to $e^x / \sqrt{x}$, as stated here (page 29).
But I obtain (knowing that $J_{\nu} (ix) \to \sqrt{\frac{2}{\pi i x}} (\cos(ix + \alpha))$ for $x \gg 1$)
$$I_{\nu}(x) \to e^{-i \frac{\pi}{2} \nu} \sqrt{\frac{2}{\pi i x}} (\cos(ix + \alpha)) = e^{-i \frac{\pi}{2} \nu} \sqrt{\frac{2}{\pi i x}} \frac{e^{-x}e^{-i\frac{\pi \nu}{2}} e^{-i \frac{\pi}{4}} + e^{x}e^{i\frac{\pi \nu}{2}} e^{i \frac{\pi}{4}}}{2}$$
where $\alpha = -\frac{\mu \pi}{2} - \frac{\pi}{4}$. This expression does not resemble $e^x / \sqrt{x}$ at all. 
2) Where is the error and how to obtain the correct result?
 A: Basically, the answer is no, because the behaviours of $I_{\nu}$ and $J_{\nu}$ are "the same" at $z=0$ (both look like $z^{\nu}$, up to a constant factor), whereas $N_{\nu}$ (or $Y_{\nu}$, as it is often called) looks like $z^{-\nu}$ when $\nu$ is not an integer, and $z^{-\nu}\log{z}$ when it is.
Another definition for $K_{\nu}(z)$ is
$$ K_{\nu}(z) = \tfrac{1}{2}\pi\frac{I_{-\nu}(z)-I_{\nu}(z)}{\sin{\nu \pi}}, $$
which is another way of seeing the asymmetry: contrast this with the usual definition of $Y_{\nu}$:
$$ Y_{\nu}(z) = \frac{J_{\nu}(z)\cos{\nu \pi}-J_{-\nu}(z)}{\sin{\nu \pi}}. $$
I would say that the asymmetry is expected because we are asking different things of each solution: $K_{\nu}$ is the only one (up to scaling) that decays at infinity, whereas $I_{\nu}$ is designed to be the one that looks like $z^{\nu}$ at zero. Of course there has to be a relation between them, since $I_{\nu},I_{-\nu},K_{\nu}$ solve the same second-order DE in the same region, but there's no reason to expect that to be symmetric.
(Compare with a simpler differential equation: $-y''+y=0$. The solution that decays at $\infty$ is $e^{-x}$, but the solution that looks like $z$ at zero is $\sinh{z}$, for example.)

For the second part of your question, remember that $e^{-x}$ decays faster than any power of $x$, so the first term in the numerator is not present. Removing that term, you find that everything cancels to the form you want.
