general relativity - Do all the spacelike curve terminate at the spatial infinity $i_0$ in the Penrose Diagram of a Schwarzchild black hole?

Let's restrict to the radial direction, so the metric can be expressed as

$ds^2=-(1-r_S/r)dt^2+(1-r_S/r)^{-1}dr^2$

with $r_S$ the Schwarzchild radius. Expressed in Kruskal coordinates, the metric is

$ds^2=4(r_S^3/r)e^{-r/r_S}(-dT^2+dR^2)$

Further coordinate transformation brings the infinity into finite coordinate values:

$\tan\frac{\tau+\rho}{2}=T+R=\sqrt{r/r_S-1}e^{(r+t)/2r_S}$

$\tan\frac{\tau-\rho}{2}=T+R=\sqrt{r/r_S-1}e^{(r-t)/2r_S}$

You can verify that the ranges of $\tau$ and $\rho$ are

$\tau\in(-\pi/2,\pi/2), \rho\in(-\pi,\pi)$

As we know that, the spatial infinity is defined at two points: $\tau=0, \rho=\pm\pi$. Now,I want to verify that all spacelike curves ($t(r),r$) parametrized by $r$ terminate at the spatial infinity and finally, I end up with counterexamples.

Because the curve $(t,r)$ is spacelike, its length is positive,

$-(1-r_S/r)(\frac{dt}{dr})^2+(1-r_S/r)^{-1}>0\Rightarrow \frac{dt}{dr}<(1-r_S/r)^{-1}$

only considering a outgoing spacelike curve outside of the horizon. So

$dt<\frac{dr}{1-r_S/r}=dr_*$

Here, $r_*$ is the tortoise coordinate. We can assume $dt=dr_*+f(r)dr$. $f(r)$ should be positive definite and not a constant function. Now, in terms of $f(r)$, we have

$\tan\frac{\tau+\rho}{2}=(r/r_S-1)e^{r/r_S}\exp[r_S^{-1}\int^r_{r_0} f(r')dr']$

$\tan\frac{\tau-\rho}{2}=-\exp[-r_S^{-1}\int^r_{r_0} f(r')dr']$

We can thus consider the behavior of the above express at $r\rightarrow\infty$. Defintely, the first expression is $+\infty$, but since $\int^r_{r_0} f(r')dr'$ is not necessarily infinite (for example, $f(r)=1/r^2$), the second expression is not necessarily $-\infty$.

Therefore, we cannot conclude that all spacelike curves terminate at the spatial infinity $i_0$ in the case of a Schwarzchild spacetime.

But all the books state the opposite. What is wrong with my calculation?