$\dot\sigma_r(t) \geq \dot\sigma_r(t_0) > 0$ for $t > t_0$. Upon integration we find that $\sigma_r(t) \geq \dot\sigma_r(t_0) (t - t_0) + \sigma_r(t_0)$.

We now find a candidate base point for a normal geodesic asymptotic to $\sigma$.

Let $\sigma_Y \colon [0, \infty) {\rightarrow}Y$ be the projection of $\sigma$ onto $Y$. Then $len( \sigma_Y ) < \infty$.

Since $\sigma$ is parametrized by arc length, $ \dot\sigma^{\alpha} \dot\sigma^{\beta} g_{\alpha \beta}(\sigma(t)) = |\dot\sigma_Y|_g \leq 1$. The metric estimate of Theorem \[thm:comparison\] implies $$L_3 \; \cosh^2( a( \sigma_r(t)+ L_1 ) ) \sum_{\alpha} (\dot\sigma_Y^{\alpha})^2 \leq \dot\sigma^{\alpha} \dot\sigma^{\beta} g_{\alpha \beta}( \sigma(t) ) \leq 1.$$ We may also consider the projected curve $\sigma_Y$. Here the metric estimates imply $$\begin{aligned}
\dot\sigma^{\alpha} \dot\sigma^{\beta} g_{\alpha \beta}( \sigma_Y(t),0 ) \leq 4 L_4 \; \sinh^2( b L_2 )  \sum_{\alpha} (\dot\sigma_Y^{\alpha})^2.\end{aligned}$$ Combining these estimates and the result of Lemma \[lemma:geodesicrgrowth\] we obtain: $$\begin{aligned}
\dot\sigma^{\alpha} \dot\sigma^{\beta} g_{\alpha \beta}( \sigma_Y(t) ) \leq  \frac{C(L_2, L_3, L_4)}{ \cosh^2( a(Ct + B) + L_1)}\end{aligned}$$ Integrating the square root of both sides we find that $len( \sigma_Y ) < \infty$.

Since $len( \sigma_Y ) < \infty$, the completeness of $M$ implies that $\sigma_Y$ has a limit, and since $Y$ is closed this limit is a point $p \in Y$. Let 