On $\text{ETH}$ with $m$ as parameter: consequences of algorithm running in time $2^{\delta m}$ where $\delta \to 0$ as $k \to \infty$

Question

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 $k \to \infty$?

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1. Which Problems Have Strongly Exponential Complexity? by R. Impagliazzo, R. Paturi and F. Zane (December 1999)
2. 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.

Answer 1

This is an answer to the updated question (the original question seems harder).

Let $\mu'_k$ be the smallest constant such that $k$-SAT that has clauses of length exactly $k$ and no trivial clauses has a $O(2^{\mu'_k m})$ time algorithm. Let $\mu_k$ be the smallest constant such that $k$-SAT with any clauses of length at most $k$ has an $O(2^{\mu_k m})$ time algorithm. I will prove that $\mu'_k = \mu_k$.

Proof: Clearly we have $\mu'_k \le \mu_k$. To prove that $\mu_k \le \mu'_k$ suppose we have an algorithm for $k$-SAT with nontrivial clauses of length exactly $k$ with time complexity $O(2^{\mu'_k m})$. To solve general $k$-SAT with $m$ clauses, we add $k$ extra variables to the input formula and $2^k-1$ extra clauses over these new variables to forbid all assignments except the all-false assignment. Now, these $k$ extra variables can be freely added to any clauses of length less than $k$ to make them length exactly $k$. Now we have a restricted instance with $m + 2^k-1$ clauses, and solve it in $O(2^{\mu'_k (m + 2^k-1)})$ time, which for constant $k$ is $O(2^{\mu'_k m})$ time.$\square$

It is easy to see that for $a \le b$ it holds that $\mu_a \le \mu_b$, and it is well-known (e.g. in section 14.1 of the book Parameterized Algorithms) that assuming ETH, $\mu_3 > 0$. Therefore showing that $\mu'_k \rightarrow 0$ as $k \rightarrow \infty$ would break ETH.