If $x(t)$ satisfy $t^2x''+tx'+(t^2-3)x=0$ then what is the limit $$ \lim_{t\rightarrow 0}\frac{x(t)}{t^{\sqrt{3}}}$$
It is very important to me, hint or full help, please. I know that the solution to the differential equation is a Bessel function so its solution are determinate by $$\{ J_\sqrt{3}(t)\ ;\ J_{-\sqrt{3}}(t)\}\ \ \ \ (because \ \alpha\not\in\mathbb{Z})$$ but I have big problem when I work with $J$ bessel. I need help please.
In the Digital Library of Mathematical Functions, you can find a lot of information on the behaviour of Bessel functions. For example, using equation (10.2.2) therein, you see that $\frac{J_{\sqrt{3}}(t)}{t^{\sqrt{3}}}$ converges as $t \downarrow 0$. For the Bessel function of the second kind, $Y_{\sqrt{3}}(t)$, you can use equation (10.2.3) in combination with equation (10.2.2) to figure out if $\frac{Y_{\sqrt{3}}(t)}{t^{\sqrt{3}}}$ converges as $t \downarrow 0$.