Prove: $\lim_{x \to 0} \frac{J_1(x)}{x}=\frac{1}{2}$
I found it to be $\frac{1}{3}$, not $\frac{1}{2}$.
My attempt:
I used this formula as an approximation (from hyperphysics)
Therefore,
$$\lim_{x \to 0} \frac{J_1(x)}{x}= \frac{\frac{(x)^1}{1 \cdot 3}}{x} = \frac{1}{3}$$
I'm guessing my error is in using an approximation?
The Bessel function of the first kind, order $1$ is denoted $J_1$ (i.e., upper case $J$) while the spherical Bessel function of the first kind, order $1$ is denoted $j_1$ (i.e., lower case $j$).
The Taylor series for $J_1(x)$ is given by
$$J_1(x)=\sum_{n=0}^\infty\frac{(-1)^n}{(n+1)!}\left(\frac x2\right)^{2n+1}$$
which has leading term $x/2$. Therefore, $\frac{J_1(x)}{x}\sim \frac12$ as $x\to 0$.
The spherical Bessel function, $j_1$, is given by $j_1(x)=\frac{\sin(x)}{x^2}-\frac{\cos(x)}{x}$. Noting the
$$\lim_{x\to 0}\frac{\frac{\sin(x)}{x^2}-\frac{\cos(x)}{x}}{x}=\frac13$$
we see that $\frac{j_1(x)}{x}\sim \frac13$ for $x\to 0$.