$$1 - \dot\sigma_r(t) \leq (1 - \dot\sigma_r(t))(1 + \dot\sigma_r(t)) = |\dot\sigma_Y|^2,$$ for large $t$. Just as in the proof of Lemma \[lemma:geodesicrgrowth\], the special form of the $r$-component of the geodesic equation in Fermi coordinates and the estimates of Theorem \[thm:comparison\] imply that $\ddot\sigma_r \geq 0$ and $$|\dot\sigma_Y|^2 \leq (1/a) \coth(a(r + L_1) \ddot\sigma_r.$$ The fundamental theorem of calculus applied to $\ddot\sigma_r$ implies that $\ddot\sigma_r \in L^1(t_0, \infty)$ and consequently since $\coth(a(r + L_1)$ is bounded, shows that $|\dot\sigma_Y|^2 \leq C \ddot\sigma_r$ for large enough $t$. Thus $|\dot\sigma_Y|^2$ is integrable and the quantity $|t-\sigma_r(t)|$ remains bounded as $t {\rightarrow}+\infty$.

*Step 2.* *$d( (p, \sigma_r(t)), (\sigma_Y(t), \sigma_r(t) ))$ is bounded as $t {\rightarrow}+\infty$:* For each $t_0$ consider the curve $$\tau^{(t_0)}(s) = ( \sigma_Y(s), \sigma_r(t_0) ), \mbox{ on} \; [t_0, \infty).$$ Clearly $ d( ( \sigma_Y(t_0), \sigma_r(t_0) ), (p, \sigma_r(t_0)) ) \leq len( \tau^{(t_0)}) $, and so it suffices to show that $len( \tau^{(t_0)})$ is bounded above by a constant independent of $t_0$. To this end we use Jacobi field estimates based at the $r$-level set of value $\sigma_r(t_0)$.

In particular consider the Jacobi field along $\gamma_{\sigma_Y(s)}(t)$ given in Fermi coordinates as the constant vector field $$J_s(t) = (\dot\sigma_Y(s), t) \in T_{(\sigma_Y(s), 