Regarding the last step of the proof of the lower bound of comparison based sorting, there is the following implication for the runtime T in my book:$T \ge n \log_2 n - n \log_2 e \implies T \in \Omega(n \log_2 n)$I tried to verify this by proof, is my conclusion correct?For any real value $x, a$ element of the extended real numbers:$f \in \Omega(g) \implies \liminf_{x \rightarrow a} \Big \lvert \frac{f(x)}{g(x)} \Big \rvert > 0$For $n \log_2 n - n \log_2 e \in \Omega(n \log_2 n)$ it has to be shown that $a$ exists so that$\liminf_{x \rightarrow a} \bigg \lvert \frac{n \log_2 n - n \log_2 e}{n \log_2 n} \bigg \rvert $ is positive.Since $\bigg \lvert \frac{n \log_2 n - n \log_2 e}{n \log_2 n} \bigg \rvert = \bigg \lvert 1 - \frac{1}{\ln n} \bigg \rvert$, the above limes inferior will be 1 for $a = \infty$.It follows that $n \log_2 n - n \log_2 e \in \Omega(n \log_2 n)$.
Proof that $n \log_2 n - n \log_2 e \in \Omega(n \log_2 n)$ (lower bound of sorting algorithms)
asymptotics