It has been shown in [1] that $k\text{-SAT}$ has a $2^{o(n)}$ algorithm if and only if it has a $2^{o(m)}$ algorithm, $n$ being the number of variables and $m$ being the number of clauses.
Being $s_k=\text{inf}\{\delta: \text{there exists }2^{\delta n}\text{ algorithm for solving }k\text{-SAT}\}$, it has been shown in [2] that, assuming $\text{ETH}$, $s_k$ increases infinitely often as $k\to \infty$, that is to say the complexity of $k\text{-SAT}$ grows as $k$ grows.
Question
Is the second statement known to hold, under $\text{ETH}$, also in the $2^{o(m)}$ realm?
More precisely, being $\mu_k=\text{inf}\{\delta: \text{there exists }2^{\delta m}\text{ algorithm for solving }k\text{-SAT}\}$, does $\text{ETH}$ imply that $\mu_k$ increases infinitely often as $k \to \infty$?
- Which Problems Have Strongly Exponential Complexity? by R. Impagliazzo, R. Paturi and F. Zane (December 1999)
- On the Complexity of $k$-SAT by R. Impagliazzo and R. Paturi (January 2001)
Update 22/11/2020
To avoid ambiguities or misunderstandings, let me clarify that here the number of clauses is not super-linear in the number of variables. The $k\text{-SAT}$ instance I'm talking about is already sparse, that is to say $m = \Theta(n)$. I'm interested in knowing the consequences on $\text{ETH}$, if any, of an algorithm running in time $2^{\delta m}$ where $\delta \to 0$ as $k \to \infty$. The existence of such algorithm proves that the sequence $\mu_k$ as defined above decreases monotonically, rather than increasing infinitely often. That's precisely why I've asked this question in the first place.
Update 26/11/2020
Following daniello's comment below, let me clarify that here each clause has exactly $k$ distinct literals, and that trivial clauses are not allowed to be present.