I‘m trying to proof that for the Bessel function of the second kind, we have that $$ \lim_{x\to 0} Y_\nu(x)= - \infty.$$
For this purpose, I want to show that for $\nu>0$ $$\lim_{x \to 0} J_v(x) ln(x)=0.$$
I know that $\lim_{x \to 0} J_v(x)=0$ for $\nu>0$ and $\lim_{x \to 0} ln(x) = -\infty.$ This led me to experimenting with L‘Hopital‘s rule but I‘ve failed to actually show the result yet. Any hints and support are very much appreciated!
Hint
Close to $x=0$ $$J_{\nu }(x)=x^n \left(\frac{2^{-\nu}}{\Gamma (\nu+1)}+O\left(x^2\right)\right)$$ $$J_{\nu }(x)\, \log(x)=\color{red}{x\,\log(x)}\,\, \,x^{\nu-1} \left(\frac{2^{-\nu}}{\Gamma (\nu+1)}+O\left(x^2\right)\right)$$ Now consider what happens to the black part of the rhs as a function of $\nu$.