I am unsure how to show this.
Suppose $\delta(s)$ defined on $(-\infty , s_*)$ is increasing and satisfies $\lim _{s\rightarrow s_*} \delta = \lim _{s\rightarrow s_*} \frac{d \delta}{d s} = \infty$ then
$e^{-\delta (s)}\int _{s_0}^se^{\delta (s')}ds'\rightarrow 0$ as $s\rightarrow s_*$ for any $s_0<s_*$.
The hint is to rewrite the integration variables with respect to $\delta$. This means we need consider
$\lim _{\delta _s\rightarrow\infty}e^{-\delta}\int _{\delta _0}^{\delta _s}e^{\delta}(\frac{d\delta}{ds})^{-1}d\delta$ but I am unable to proceed further.
Any help is much appreciated.
Why don't you use De l'Hospital? $$ \lim_{s \to s^*} \frac{e^{\delta(s)}}{e^{\delta(s)}\delta'(s)}=0. $$