In 1.7 on p.224 of the following paper, there is a rigidity result for compact manifolds whose sectional curvature is almost $-1$.

[Gromov, M.. Manifolds of negative curvature. J. Differential Geometry 13 (1978), no. 2, 223-230.][1]


**Statement:** Given $n \geq 4$ and $C>0$, there is an $\epsilon_0=\epsilon_0(n, C)>0$ so that
for any $0<\epsilon<\epsilon_0$, if a compact
Riemannian manifold $(M^n,g)$ satisfies $-1 \leq K \leq -1+\epsilon$ and the volume $Vol(M,g) \leq C$, then it must admit a metric of constant negative sectional curvature.

The above result is mentioned as a standard application from the main "non-collapsing" result (in Sec 1.2 on p.223 of the above paper). Roughly speaking, under the above assumption the volume has a lower bound, and the diameter has an upper bound. However I could not figure out the proof of Statement.

Assume by contradiction, we have a sequence of $(M_i, g_i)$ whose curvature is $-1<K<-1+\epsilon_i$ and $Vol(M_i, g_i) \leq C$, but each of $M_i$ does not admit a hyperbolic metric. then we may apply the $C^{1, \alpha}$ convergence theory of Cheeger-Gromov. Now we get a smooth manifold $N$ and diffeomorphisms $f_i: N \rightarrow M_i$ so that the pull back $f_i^{\ast} g_i$ converges to a $C^{1, \alpha}$ metric $h$ on $N$ under the $C^{1, \alpha}$ norm. Even if $\alpha$ could be $1$, the convergence is still weaker than $C^2$.

**Question:** Why is $(N, h)$ a hyperbolic metric? 


  [1]: https://projecteuclid.org/journals/journal-of-differential-geometry/volume-13/issue-2/Manifolds-of-negative-curvature/10.4310/jdg/1214434487.full