Limit involving ratios of inverse functions

Question

Let $H(t)$ be an strictly increasing function such that $\lim_{t\to \infty}H(t)=\infty$. Assuming that $$\lim_{t\to \infty}\frac{H(t)}{\log t}=1$$ I would like to prove or disprove that $$\lim_{y\to \infty}\frac{H^{-1}(y)}{\exp(y)}=1$$ I have not been able to find a counterexample but also not been able to prove if it is true! Looks to me this has to be true!

Answer 1

A counterexample. Take $H(t)=\ln(t)+\ln(\ln(t))$ then $$\lim_{t\to +\infty}\frac{H(t)}{\ln(t)} =1.$$ Moreover, by letting $y=H(t)$, we have that $$\lim_{y\to +\infty}\frac{H^{-1}(y)}{\exp(y)}=\lim_{t\to +\infty}\frac{t}{\exp(H(t))}=\lim_{t\to +\infty}\frac{t}{\exp(\ln(t)+\ln(\ln(t)))}\\=\lim_{t\to +\infty}\frac{t}{t\cdot\ln(t)}=\lim_{t\to +\infty}\frac{1}{\ln(t)}=0.$$