Let $\varphi(t)$ be a (strictly) increasing concave function with $\varphi(0)=0$, and $B>1$. It´s easy to see that \begin{equation} 1 \le \liminf_{ t\to 0^+} \frac{\varphi(Bt)}{\varphi(t)} \le B. \end{equation} We are looking for an example that fullfill the hypotesys and where the limit is exactly $1$.
This type of limits appears, for example, to determine whether the Orlicz space of sequences $\ell_{\varphi}$ is locally bounded or not (the answer is yes, iff the limit is greater than $1$ for some constant $B>1$).
Our original setup is taking $\varphi(t)=\int_0^t \phi$, where $\phi$ is positive and decreasing (non-increasing), integrable in $(0,\alpha)$ for some $\alpha >0$. In this context, if $\phi$ is bounded then the limit is always $B$. So, we have to look an unbounded functión. Taking $\phi(t)=t^{-r}, 0<r<1$, the limit is greater than 1. For $r=1$ the limit is (would be, using L´Hospital rule) $1$, but $\varphi$ is no longer defined, as the function $1/t$ is not integrale in $0$.
Is there an example where de limit is exactly $1$? Futher, is there a way to characterize the class of functions $\varphi$ for which the limit is $1$?
Define $$ \varphi(t) = \begin{cases} 0 & \text{ if } t = 0 \, ,\\ -1/\ln(t) & \text{ if } 0 < t < e^{-2} \, . \end{cases} $$ It can be verified easily that $\varphi $ is strictly increasing and stricly concave on $[0, e^{-2}]$. For every $B > 1$ is $$ \lim_{t \to 0^+}\frac{\varphi(Bt)}{\varphi(t)} = \lim_{t \to 0^+}\frac{\ln(t)}{\ln(t) + \ln(B)} = 1 \, . $$
(If necessary, $\varphi$ can be extended to an increasing concave function on $[0, \infty)$, this does not affect the limit of the quotient at $t = 0$.)
Remark: For a concave increasing function $\varphi$ with $\varphi(0) = 0$ is $\varphi(t)/t$ decreasing, so that the limit $$ \lim_{t \to 0^+}\frac{\varphi(t)}{t} $$ exists, but can be $+\infty$. If that limit is finite then $$ \lim_{t \to 0^+}\frac{\varphi(Bt)}{\varphi(t)} = B $$ follows, therefore the idea was to look for functions with a vertical asymptote at the origin.