How prove this $\displaystyle\lim_{\tau\to t}f(t,\tau)=\frac{1}{2\pi}\frac{x'_{1}(t)x''_{2}(t)-x'_{2}(t)x''_{1}(t)}{[x'_{1}(t)]^2+[x'_{2}(t)]^2}$ Question:
let 
$$j_{1}(t)=\sum_{p=0}^{\infty}\dfrac{(-1)^p}{p!(1+p)!}\left(\dfrac{t}{2}\right)^{1+2p}$$
$$Y_{1}(t)=\dfrac{2}{\pi}\left(\ln{\dfrac{t}{2}}+C\right)j_{1}(t)-\dfrac{1}{\pi}\sum_{p=0}^{\infty}\dfrac{(-1)^p}{p!(1+p)!}\left(\dfrac{t}{2}\right)^{1+2p}\left(\sum_{m=1}^{p+1}\dfrac{1}{m}+\sum_{m=1}^{p}\dfrac{1}{m}\right)\tag{1}$$
let
$$f(t,\tau)=\dfrac{ik}{2}\left(x'_{2}(\tau)[x_{1}(\tau)-x_{1}(t)]-x'_{1}(\tau)[x_{2}(\tau)-x_{2}(t)]\right)\dfrac{j_{1}(k|x(t)-x(\tau)|)+iY_{1}(k|x(t)-x(\tau)|)}{|x(t)-x(\tau)|}$$
where $C$ is Euler constant,and
$$|x(t)-x(\tau)|=\sqrt{[x_{1}(t)-x_{1}(\tau)]^2+[x_{2}(t)-x_{2}(\tau)]^2}$$

show that:
  $$\lim_{\tau\to t}f(t,\tau)=\frac{1}{2\pi}\dfrac{x'_{1}(t)x''_{2}(t)-x'_{2}(t)x''_{1}(t)}{[x'_{1}(t)]^2+[x'_{2}(t)]^2}$$

This problem is from (Inverse Acoustic and Electromagnetic Scattering Theory ) page 77,this author can't post his solution,and I take sometime to prove this,But I can't prove it.
Now  my try: since
$$J_{1}(t)\approx \dfrac{t}{2}+o(t)$$
$$Y_{1}(t)\approx\dfrac{t}{\pi}\left(\ln{\dfrac{t}{2}}+C\right)-\dfrac{t}{2\pi}\tag{2}$$
and then for $f(t,\tau)$ I want use L'Hôpital's rule
Thank you  for you help! Thank you Thank you
 A: Observe that $J_1(t)$ and $Y_1(t)$ are the Bessel function of the first kind and the second kind respectively, and $H_1^{(1)}(t)=J_1(t)+iY_1(t)$ is the Hankel function of first kind. For small argument we have
$$
\begin{align}
J_1(t)&\sim \frac{t}{2}\tag 1\\
Y_1(t)&\sim \frac{2}{\pi}\left[\log\left(\frac{t}{2}+C\right)\right]\frac{t}{2}-\frac{2}{\pi t}\tag 2
\end{align}
$$
so that 
$$
H_1^{(1)}(t)=J_1(t)+iY_1(t)\sim -\frac{2i}{\pi t}.\tag 3
$$
Let's call 
$$
\begin{align}
\psi(\tau,t)&=x'_{2}(\tau)[x_{1}(\tau)-x_{1}(t)]-x'_{1}(\tau)[x_{2}(\tau)-x_{2}(t)]\\
r(\tau,t)&=|x(t)-x(\tau)|=\sqrt{[x_{1}(t)-x_{1}(\tau)]^2+[x_{2}(t)-x_{2}(\tau)]^2}
\end{align}
$$
so that
$$
f(\tau,t)=\frac{ik}{2}\psi(\tau,t)\frac{H_1^{(1)}(kr)}{r}.
$$
For $\tau\to t$ so that $r(\tau,t)\to 0$ we can use (3) and then
$$
\frac{ik}{2}\frac{H_1^{(1)}(kr)}{r}\sim \frac{1}{\pi r^2}
$$
and for $f(\tau,t)$
$$
f(\tau,t)\sim \frac{\psi(\tau,t)}{\pi r^2(\tau,t)}.
$$
Put, for $k=1,2$
$$
\xi_k(\tau,t)=\frac{x_k(\tau)-x_k(t)}{\tau-t}
$$
and observe that for $\tau\to t$ we have $\xi_k(\tau,t)\to x'_k(t)$.
We can write
$$
f(\tau,t)\sim \frac{\psi(\tau,t)}{\pi r^2(\tau,t)}=\frac{1}{\pi} \frac{\varphi(\tau,t)}{\xi_1^2(\tau,t)+\xi_2^2(\tau,t)}\to \frac{1}{\pi} \frac{\varphi(t,t)}{[x'_{1}(t)]^2+[x'_{2}(t)]^2}\tag 4
$$
with
$$
\varphi(t,t)=\lim_{\tau\to t}\varphi(\tau,t)=\frac{x_2'(\tau)[x_1(\tau)-x_1(t)]-x_1'(\tau)[x_2(\tau)-x_2(t)]}{(\tau-t)^2}.
$$
Using the l'Hôpital's rule we find
$$
\begin{align}
\lim_{\tau\to t}\varphi(\tau,t)&=\lim_{\tau\to t}\tfrac{x_2''(\tau)[x_1(\tau)-x_1(t)] +x_2'(\tau)x_1'(\tau)-x_1''(\tau)[x_2(\tau)-x_2(t)]-x_1'(\tau)x_2'(\tau)}{2(\tau-t)}\\
&=\lim_{\tau\to t}\frac{1}{2}\left(x_2''(\tau)\xi_1(\tau,t)-x_1''(\tau)\xi_2(\tau,t)\right)\\
&=\frac{1}{2}\left(x_2''(t)x_1'(t)-x_1''(t)x_2'(t)\right).\tag 5
\end{align}
$$
Then from (4) and (5) we have

$$
\lim_{\tau\to t}f(\tau,t)=\frac{1}{2\pi}\dfrac{x'_{1}(t)x''_{2}(t)-x'_{2}(t)x''_{1}(t)}{[x'_{1}(t)]^2+[x'_{2}(t)]^2}.
$$

