I am trying to find the limit: $lim_{x \rightarrow 0} x\frac{I_{\nu}'(x)}{I_{\nu}(x)}$ for real order $\nu > 0$.
Wolfram Alpha gives the result of $\nu$ for real order $\nu > 0$, but I can't see why. The recurrence relations regarding derivatives $2 I'_{\nu}(x) = I_{\nu-1}(x) + I_{\nu+1}(x)$ doesn't seem the right thing to use.
Any suggestions?
Use the alternative formula http://dlmf.nist.gov/10.29.E2 $$I_{\nu}'(x)= I_{\nu+1}+\frac{\nu}{x}I_{\nu}(x)$$ and the asymptotic form for $I_{\nu}(x)\sim(\frac{x}{2})^{\nu}/\Gamma(\nu+1)\;$ for $x\rightarrow 0.$ Then you get $$x\frac{I_{\nu}'(x)}{I_{\nu}(x)} =x\left(\frac{I_{\nu+1}(x)}{I_{\nu}(x)} + \frac{\nu}{x}\frac{I_{\nu}(x)}{I_{\nu}(x)}\right) \sim\nu + x\frac{(\frac{x}{2})^{\nu+1}\Gamma(\nu+1)}{(\frac{x}{2})^{\nu}\Gamma(\nu+2)}\\ =\nu + x\frac{(\frac{x}{2})}{\nu+1}\\ $$ and therefore your limit is $$\lim_{x\rightarrow 0} x\frac{I_{\nu}'(x)}{I_{\nu}(x)}=\nu$$