Let us consider the sum $$m_N(x)=\frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{cN}{2}-i(2c-1)\right).$$ The first step to get an asymptotic approximation is to extract the leading term in $N$ to obtain $$m_N(x)=[f(x)]\log N+\frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-\frac{i}{N}(2c-1)\right).$$ When $N$ is finite, we recognize a Riemann series and apply the average theorem. So, there exists a value of argument of the logarithm such that $$m_N(x)=[f(x)]\log N+[f(x)]\log[z(x)].$$ We can take $z(x)=\frac{c}{2}-t[f(x)](2c-1)$ being $t\in (0,1)$.
Indeed, we can define a partition with $x_i=x_{i-1}+\frac{1}{[f(x)]N}$ and so $$\frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-\frac{i}{N}(2c-1)\right)=[f(x)]\Delta x\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-i[f(x)](2c-1)\Delta x\right)$$ being $\Delta x=\frac{1}{[f(x)]N}$. But this, in the given limit, is nothing else than $$\int_{\frac{c}{2}}^{\frac{c}{2}-[f(x)](2c-1)}\log(x)dxint_{\frac{c}{2}}^{\frac{c}{2}-[f(x)](2c-1)}\log(z)dz<\infty$$ as it should.
Let us consider the sum $$m_N(x)=\frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{cN}{2}-i(2c-1)\right).$$ The first step to get an asymptotic approximation is to extract the leading term in $N$ to obtain $$m_N(x)=[f(x)]\log N+\frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-\frac{i}{N}(2c-1)\right).$$ When $N$ is finite, we recognize a Riemann series and apply the average theorem. So, there exists a value of argument of the logarithm such that $$m_N(x)=[f(x)]\log N+[f(x)]\log[z(x)].$$ We can take $z(x)=\frac{c}{2}-t[f(x)](2c-1)$ being $t\in (0,1)$.
Indeed, we can define a partition with $x_i=x_{i-1}+\frac{1}{[f(x)]N}$ and so $$\frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-\frac{i}{N}(2c-1)\right)=[f(x)]\Delta x\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-i[f(x)](2c-1)\Delta x\right)$$ being $\Delta x=\frac{1}{[f(x)]N}$. But this, in the given limit, is nothing else than $$\int_{\frac{c}{2}}^{\frac{c}{2}-[f(x)](2c-1)}\log(x)dx<\infty$$ as it should.